BCA-504

OPERATIONS RESEARCH BCA - 504

OPERATIONS RESEARCH BCA - 504

This SIM has been prepared exclusively under the guidance of Punjab Technical University (PTU) and reviewed by experts and approved by the concerned statutory Board of Studies (BOS). It conforms to the syllabi and contents as approved by the BOS of PTU.

Reviewer Dr. N. Ch. S.N. Iyengar

Senior Professor, School of Computing Sciences, VIT University, Vellore

Authors: C R Kothari (Units: 1.0-1.2, 1.4-1.11) Copyright © C R Kothari, 2008 S Kalavathy (Units: 1.3, unit 2, 3, 4) Copyright © Reserved, 2008 Reprint 2010 All rights reserved. No part of this publication which is material protected by this copyright notice may be reproduced or transmitted or utilized or stored in any form or by any means now known or hereinafter invented, electronic, digital or mechanical, including photocopying, scanning, recording or by any information storage or retrieval system, without prior written permission from the Publisher. Information contained in this book has been published by VIKAS® Publishing House Pvt. Ltd. and has been obtained by its Authors from sources believed to be reliable and are correct to the best of their knowledge. However, the Publisher and its Authors shall in no event be liable for any errors, omissions or damages arising out of use of this information and specifically disclaim any implied warranties or merchantability or fitness for any particular use.

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CAREER OPPORTUNITIES

The significance of the role played by Operations Research in the growth of any corporate or business entity cannot be overemphasized, as it forms a substantial part of the all-important quantitative methods employed by business concerns in decision making. Thus, students specializing in Operations Research stand to avail opportunities in the Production and Operations areas, in particular, and practically all the other departments including Finance, as a fair amount of knowledge of OR is required in all these areas.

A major source of employment is obviously the manufacturing industry where linear programming and quantitative techniques are employed while taking all decisions—big or small. Further, their capabilities and aptitude would be well—received where Logistics and Transportation are involved. In fact, the field of OR is indispensable for any concern. Thus, even in the hospitality industry there will be ample career opportunities for them. And, of course, academics and research are other areas where people with sound OR knowledge would be welcome.

PTU DEP SYLLABI-BOOK MAPPING TABLE BCA - 504 Operations Research

Section I Origin & Development of OR, Nature & Characteristics Features of OR, Models & Modeling in Operation Research. Methodology of OR, General Methods for Solving OR Models, OR & Decision Making, Application, Use & Limitations of OR. Section - II Linear Programming: Formulation, Graphical, Big M Method & Simplex Method, Duality in L.P.: Conversion of Primal to Dual Only. Transportation Problems: Test for Optimality, Degeneracy in Transportation Problems. Unbalanced Transportation, Assignment Problems, Traveling Salesman Problem. Section - III Decision Making: Decision Making Environment, Decision under Uncertainty, Decision under Risk, Decision Tree Analysis. Integer Programming and Dynamic Programming: Concept and Advantages only.

Unit 1: Introduction to Operations Research (Pages 3-24)

Unit 2: Linear Programming (Pages 25-75); Unit 3: Transportation Problems (Pages 77-131)

Unit 4: Decision-Making (Pages 133-163)

CONTENTS INTRODUCTION UNIT 1

INTRODUCTION TO OPERATIONS RESEARCH

1 3-24

1.0 Introduction 1.1 Unit Objectives 1.2 Origin and Development of Operations Research 1.2.1 Nature and Characteristic Features of Operations Research (OR)

1.3 Models and Modelling in Operations Research 1.3.1 Advantages of a Model 1.3.2 Classification of Models

1.4 Methodology of Operations Research 1.5 General Methods for Solving OR Models 1.5.1 Operations Research and Decision-Making

1.6 Applications and Limitations of OR 1.6.1 1.6.2 1.6.3 1.6.4

1.7 1.8 1.9 1.10 1.11

Applications of OR OR and Modern Business Management Limitations of OR Froniters/Scope of OR

Summary Key Terms Answers to ‘Check Your Progress’ Questions and Exercises Further Reading

UNIT 2 LINEAR PROGRAMMING 2.0 Introduction 2.1 Unit Objectives 2.2 Linear Programming Formulation 2.2.1 General Formulation of LPP 2.2.2 Matrix Form of LPP

2.3 Graphical Method 2.3.1 2.3.2 2.3.3 2.3.4 2.3.5

Procedure for Solving LPP by Graphical Method General Formulation of LPP Matrix Form of LPP Some Important Definitions Canonical or Standard Forms of LPP

2.4 Big M Method 2.5 Simplex Method 2.5.1 Introduction 2.5.2 Definition

2.6 Duality in LP: Conversion of Primal to Dual 2.6.1 Formulation of Dual Problem 2.6.2 Definition of Dual Problem 2.6.3 Important Results in Duality

2.7 2.8 2.9 2.10 2.11


25-75

UNIT 3 3.0 3.1 3.2 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11

TRANSPORTATION PROBLEMS

Introduction Unit Objectives Transportation Problems Test for Optimality Degeneracy in Transportation Problems Unbalanced Transportation Assignment Problems Travelling Salesman Problem Summary Key Terms Answers to ‘Check Your Progress’ Questions and Exercises Further Reading

UNIT 4 DECISION-MAKING 4.0 Introduction 4.1 Unit Objectives 4.2 Decision-Making Environment 4.2.1 Decision-Making under Certainty 4.2.2 Decision-Making under Uncertainty 4.2.3 Decision-Making under Risk

4.3 Decision Tree Analysis 4.4 Integer Programming and Dynamic Programming: Concepts and Advantages Only 4.4.1 4.4.2 4.4.3 4.4.4 4.4.5 4.4.6

4.5 4.6 4.7 4.8 4.9

77-131

Importance of Integer Programming Problems Applications of Integer Programming Decision Tree and Bell Man’s Principle of Optimality Characteristics of DPP Dynamic Programming Algorithm Advantages of DPP


133-163

Introduction

INTRODUCTION Operational Research, or simply OR, originated in the context of military operations, but today it is widely accepted as a powerful tool for planning and decision-making, especially in business and industry. The OR approach has provided a new tool for managing conventional management problems. In fact, operational research techniques do constitute a scientific methodology of analysing the problems of the business world. They provide an improved basis for taking management decisions. The practice of OR helps in tackling intricate and complex problems such as that of resource allocation, product mix, inventory management, sequencing and scheduling, replacement, and a host of similar problems of modern business and industry.

NOTES

With IT facilities becoming widely available, the significance and scope of OR has grown, and is still growing. Hence, OR is now an integral part of courses of computer science, economics, business management, public administration and several other disciplines. Keeping all this in view, the book Operations Research introduces the basic elements of OR in the simplest possible way. The book has four units. Unit 1 describes the origin and development of operations research and its characteristics, besides elaborating on various models and methodology. Unit 2 explains Linear Programming, which is a decision-making technique. The unit describes various methods of solving a LP problem. Unit 3 deals with Transportation and Assignment Problems and the methods to find optimal solutions. Unit 4 explains the importance of decision-making and the situations in which decision-making plays a crucial role. A Summary and Key Terms are provided at the end of each unit, for quick recollection. Besides, numerous examples are provided in the text. The ‘Check Your Progress’ questions and ‘Questions and Exercises’ sections will further help you in understanding the contents in a better way.

Self-Instructional Material 1

UNIT 1 INTRODUCTION TO OPERATIONS RESEARCH

Introduction to Operations Research

NOTES Structure 1.0 Introduction 1.1 Unit Objectives 1.2 Origin and Development of Operations Research 1.2.1 Nature and Characteristic Features of Operations Research (OR)

1.3 Models and Modelling in Operations Research 1.3.1 Advantages of a Model 1.3.2 Classification of Models

1.4 Methodology of Operations Research 1.5 General Methods for Solving OR Models 1.5.1 Operations Research and Decision-Making

1.6 Applications and Limitations of OR 1.6.1 1.6.2 1.6.3 1.6.4

1.7 1.8 1.9 1.10 1.11

1.0

Applications of OR OR and Modern Business Management Limitations of OR Frontiers/Scope of OR


INTRODUCTION

Modern technological advance is accompanied by a growth of scientific techniques. While existing methods have been improved to meet the challenge of problems arising from the development of commerce and industry, a large number of new techniques or so-called sophisticated tools of analysis have been and are being devised to enlarge the extent of scientific knowledge to unlimited bounds of its applications. Such techniques have brought about a virtual revolution and can be reckoned as controlling forces in different walks of life. Operations Research, popularly known as OR, is a recent addition to a long list of scientific tools which provide a new outlook to many conventional management problems. OR adds greater sophistication in solving management problems. It seeks the determination of the best (optimum) course of action of a decision problem, given the limited resources. Therefore, OR has become a versatile tool in the field of management and its potential for future use is indeed substantial. With the advent of automation, the centralization of organizational management has been disintegrated. In a single unit of industry, different departments of production, sales and inventory are governed by specialized agencies. Their targets and goals often disagree in an attempt to serve the common purpose of the organization. The policy decisions to coordinate these conflicting directives may be taken quite effectively, if an optimal solution can be determined from all available Self-Instructional Material 3

Operations Research

NOTES

alternative compromise formulae. OR provides an effective scientific technique to solve such decision-making problems of modern business and industry.

1.1

UNIT OBJECTIVES

After going through this unit, you will be able to: ●

Trace the origin and development of operations research

●

Understand the various types of models in operations research and their advantages

●

Describe the methodology of operations research

●

Understand the role of decision-making in operations research

●

Describe the application, use and limitations of operations research

1.2

ORIGIN AND DEVELOPMENT OF OPERATIONS RESEARCH

The subject of Operations Research (OR) was developed in military context during World War II, pioneered by the British scientists. At that time, the military management in England appointed a study group of scientists to deal with die strategic and tactical problems related to air and land defence of the country. The main reason for conducting the study was that they were having very limited military resources. It was, therefore, necessary to decide upon the most effective way of utilizing these resources. As the name implies, operations research was apparently invented because the team was dealing with research on (military) operations. The scientists studied the various problems and on the basis of quantitative study of operations suggested certain approaches which showed remarkable success. The encouraging results obtained by the British operations research teams consisting of personnel drawn from various fields like mathematics, physics, biology, psychology and other physical sciences, quickly motivated the United States military management to start similar activities. Successful innovations of the US teams included the development of new flight patterns, planning sea mining and effective utilization of electronic equipment. Similar OR teams also started functioning in Canada and France. These OR teams were usually assigned to the executive-incharge of operations and as such their work came to be known as ‘Operational Research’ in the UK and by a variety of names in the United States: ‘operational analysis, operations evaluation, operations research, systems analysis, systems evaluation and systems research. The name ‘operational research’ or ‘operations research’ or simply OR is most widely used nowadays all over the world, for the new approach to systematic and scientific study of the operations of the system. Till the 1950s, use of operations research was mainly confined to military purposes. After the end of World War II, the success of military teams attracted the attention of industrial managers who were seeking solutions to their complex managerial problems. At the end of the War, expenditures on defence research were reduced in the UK and this led to the release of many OR workers from the military at a time when industrial managers were confronted with the need to reconstruct most of 4 Self-Instructional Material

Britain’s manufacturing industries and plants that had been damaged in War. Executives in such industries sought assistance from the said OR workers. But in the USA, most war experienced operations research workers remained in military service as the defence research was increased and consequently, operations research was expanded at the end of the War. It was only in the early 1950s, that industry in the USA began to absorb the operations research workers under the pressure for increased demands for greater productivity originated because of the outbreak of the Korean conflict and because of technological developments in industry! Thus, operations research began to develop in industrial field in the United States since the year 1950. The Operations Research Society of America was formed in 1953 and the International Federation of Operations Research was formed in 1957.


NOTES

Societies were established. Various journals relating to operations research began to appear in different countries in the years that followed the mid-1950s. Courses and curricula in operations research in different universities and other academic institutions began to proliferate in the United States. Other countries rapidly followed suit and thus, Operations Research came to be applied for solving business and industrial problems. Introduction of Electronic Data Processing (EDP) methods further enlarged the scope for application of OR techniques. With the help of a digited computer many complex problems can be studied on a day-to-day basis. As a result, many industrial concerns are adopting OR as an integrated decision-making tool for their routine decision procedures.

1.2.1 Nature and Characteristic Features of Operations Research (OR) There have been various definitions for OR like applied decision-making, quantitative commonsense and making of economic decisions. The following definitions are generally talked about and widely accepted: 1. ‘OR is the application of scientific methods, techniques and tools to problems involving operations of systems so as to provide those in control of operations with optimum solutions to the problems’. — Churchman, Ackoff and Arnoff 2. ‘Operations Research is a scientific method of providing executive departments with a quantitative basis for decisions regarding the operations under their control’. — P.M. Morse and G.F. Kimball 3. ‘Operations Research is the attack of modern science on problems of likelihood that arise in the management and control of men and machines, materials and money in their natural environment, its special technique is to invent a strategy of control by measuring, comparing and predicting probable behaviour through a scientific model of a situation’. — Stafford Beer 4. ‘Operations Research is applied decision theory. It uses any scientific, mathematical or logical means to attempt to cope with the problems that confront the executive when he tries to achieve a thorough going rationality in dealing with his decision problems’. — D.W. Miller and M.K. Starr Self-Instructional Material 5

Operations Research

NOTES

From these definitions, we can state that Operational research can be considered as being the application of scientific method by interdisciplinary teams to problems involving the control of organized (man-machine) systems so as to provide solutions which best serve the purposes of the organization as a whole. The different characteristics constituting the nature of OR can be summed up as follows: (1) Interdisciplinary team approach: Operations research has the characteristics that it is done by a team of scientists drawn from various disciplines such as mathematics, statistics, economics, engineering, and physics. It is essentially an interdisciplinary team approach. Each member of the OR team is benefited from the viewpoint of the other, so that a workable solution obtained through such a collaborative study has a greater chance of acceptance by management. (2) Systems approach: Operations research emphasizes on the overall approach to the system. This characteristic of OR is often referred to as system orientation. The orientation is based on the observation that in the organized systems the behaviour of any part ultimately has some effect on every other part. But all these effects are not significant and even not capable of detection. Therefore, the essence of system orientation lies in the systematic search for significant interactions in evaluating actions of any part of the organization. In O.R., an attempt is made to take account of all the significant effects and to evaluate them as a whole. OR thus considers the total system for getting the optimum decisions. (3) Helpful in improving the quality of solution: Operations research cannot give perfect answers or solutions to the problems. It merely gives bad answers to the problems which otherwise have worst answers. Thus, OR simply helps in improving the quality of the solution but does not result in perfect solution. (4) Scientific method: Operations research involves scientific and systematic attack of complex problems to arrive at the optimum solution. In other words, it uses techniques of scientific research. Thus it comprehends both aspects, i.e., it includes both scientific research on the phenomena of operating systems and the associated engineering activities aimed at applying the results of research. (5) Goal-oriented optimum solution: Operations research tries to optimize a well-defined function subject to given constraints and as such is concerned with the optimization theory. (6) Use of models: Operations research uses models built by quantitative measurement of the variables concerning a given problem and also derives a solution from the model using one or more of the diversified solution techniques. A solution may be extracted from a model either by conducting experiments on it or by mathematical analysis. The purpose is to help the management to determine its policy and actions scientifically.

6 Self-Instructional Material

(7) Requires willing executives: Operations research does require the willingness on the part of the executive for experimentation to evaluate the costs and the consequences of the alternative solutions of the problem. It enables the decision-maker to be objective in choosing an alternative from among many possible alternatives.


NOTES

(8) Reduces complexity: Operations research tries to reduce the complexity of business operations and does help the executive in correcting a troublesome function and to consider innovations which are too costly and complicated to experiment with the actual practice. In view of this above, OR must be viewed both as a science and as an art. As science, OR provides mathematical techniques and algorithms for solving appropriate decision problems. OR is an art because success in all the phases that precede and succeed the solution of a problem largely depends on the creativity and personal ability of the decision-making analysts.

1.3

MODELS AND MODELLING IN OPERATIONS RESEARCH

A model in OR is a simplified representation of an operation, or is a process in which only the basic aspects or the most important features of a typical problem under investigation are considered. The objective of a model is to identify significant factors and interrelationships. The reliability of the solution obtained from a model depends on the validity of the model representing the real system. A good model must possess the following characteristics: (i) It should be capable of taking into account, new formulation without having any changes in its frame. (ii) Assumptions made in the model should be as small as possible. (iii) Variables used in the model must be less in number ensuring that it is simple and coherent. (iv) It should be open to parametric type of treatment. (v) It should not take much time in its construction for any problem.

1.3.1 Advantages of a Model There are certain significant advantages in using a model. These are: (i) Problems under consideration become controllable. (ii) It provides a logical and systematic approach to the problem. (iii) It provides the limitations and scope of an activity. (iv) It helps in finding useful tools that eliminate duplication of methods applied to solve problems. (v) It helps in finding solutions for research and improvements in a system. (vi) It provides an economic description and explanation of either the operation, or the systems it represents. Self-Instructional Material 7

Operations Research

1.3.2 Classification of Models The classification of models is a subjective problem. They may be distinguished as follows:

NOTES

(1) Models by degree of abstraction (2) Models by function (3) Models by structure (4) Models by nature of an environment (5) Models by the extent of generality

1.3.2.1 Models by function These models consist of (a) Descriptive models (b) Predictive models and (c) Normative models. Descriptive models: They describe and predict facts and relationships among the various activities of the problem. These models do not have an objective function as part of the model to evaluate decision alternatives. In these models, it is possible to get information as to how one or more factors change as a result of changes in other factors. Normative or optimization models: They are prescriptive in nature and develop objective decision-rule for optimum solutions.

1.3.2.2 Models by structure These models are represented by (a) Iconic models (b) Analogue models, and (c) Mathematical or symbolic models. Iconic or physical models: They are pictorial representations of real systems and have the appearance of the real thing. An iconic model is said to be scaled down or scaled up according to the dimensions of the model which may be smaller or greater than that of the real item, e.g., city maps, houses blueprints, globe, and so on. These models are easy to observe and describe, but are difficult to manipulate and are not very useful for the purpose of prediction. Analog models: These are more abstract than the iconic ones for there is no look alike correspondence between these models and real life items. The models in which one set of properties is used to represent another set of properties are called analog models. After the problem is solved, the solution is reinterpreted in terms of the original system. These models are less specific, less concrete, but easier to manipulate than iconic models. Mathematic or symbolic models: They are most abstract in nature. They employ a set of mathematical symbols to represent the components of the real system. These variables are related together by means of mathematical equations to describe the behaviour of the system. The solution of the problem is then obtained by applying well developed mathematical techniques to the model. The symbolic model is usually the easiest to manipulate experimentally and it is the most general and abstract. Its function is more explanatory than descriptive. 8 Self-Instructional Material

1.3.2.3 Models by nature of an environment


These models can be further classified into (a) Deterministic models and (b) Probabilistic models. Deterministic models: They are those in which all parameters and functional relationships are assumed to be known with certainty when the decision is to be made. Linear programming and break-even models are the examples of deterministic models.

NOTES

Probabilistic or stochastic models: These models are those in which atleast one parameter or decision variable is a random variable. These models reflect to some extent the complexity of the real world and the uncertainty surrounding it.

1.3.2.4 Models by the extent of generality These models can be further categorized into (a) Specific models (b) General models When a model presents a system at some specific time, it is known as a specific model. In these models, if the time factor is not considered, they are termed as static models. An inventory problem of determining economic order quantity for the next period assuming that the demand in planning period would remain same as that of today is an example of static model. Dynamic programming may be considered as an example of dynamic model. Simulation and Heuristic models fall under the category of general models. These models are used to explore alternative strategies which have been overlooked previously.

1.4

METHODOLOGY OF OPERATIONS RESEARCH

The methodology of OR generally involves the following steps: (1) Formulating the Problem. The first step in an OR study is to formulate the problem in an appropriate form. Formulating a problem consists in identifying, defining and specifying the measures of the components of a decision model. This means that all quantifiable factors which are pertinent to the functioning of the system under consideration are defined in mathematical language: variables (factors which are not controllable), and parameters or coefficient, along with the constraints on the variables and the determination of suitable measures of effectiveness. (2) Constructing the Model. The second step consists in constructing the model by which we mean that appropriate mathematical expressions are formulated which describe the interrelations of all variables and parameters. In addition, one or more equations or inequalities are required to express the fact that some or all of the controlled variables can only be manipulated within limits. Such equations or inequalities are termed as constraints or the restrictions. The model must also include an objective function which defines the measure of effectiveness of the system. The objective function and the constraints together constitute a model of the problem that we want to solve. This model describes the technology and the economics of the system under consideration through a set of simultaneous equations and inequalities.

Check Your Progress 1. Mention any one characteristic feature of operations research. 2. List any two advantages of a model. 3. How are models classified? Self-Instructional Material 9

Operations Research

NOTES

(3) Deriving the solution. Once the model is constructed the next step in an OR study is that of obtaining the solution to the model, i.e., finding the optimal values of the controlled variables—values that produce the best performance of the system for specified values of the uncontrolled variables. In other words, an optimum solution is determined on the basis of the various equations of the model satisfying the given constraints and interrelations of the system and at the same time maximizing profit or minimizing’cost or coming as close as possible to some other goal or criterion. How the solution can be derived depends on the nature of the model. In general, there are three methods available for the purpose viz., the analytical methods, the numerical methods and the simulation methods. Analytical methods involve expressions of the model by mathematical computations and the kind of mathematics required depends upon the nature of the model under consideration. This sort of mathematical analysis can be conducted only in some cases without any knowledge of the values of the variables, but in others the values of the variables must be known concretely or numerically. In latter cases, we use the numerical methods which are concerned with iterative procedures through the use of numerical computations at each step. The algorithm (or the set of computational rules) is started with a trial or initial solution and continued with a set of rules for improving it towards optimality. The initial solution is then replaced by the improved one and the process is repeated until no further improvement is possible. But in those cases where the analytical as well as the numerical methods cannot be used for deriving the solution, we use simulation methods, i.e., we conduct experiments on the model in which we select values of the uncontrolled variables with the relative frequencies dictated by their probability distributions. The simulation methods involve the use of probability and sampling concepts, and are generally used with the help of computers. Whichever method is used, our objective is to find an optimal or near-optimal solution, i.e., a solution which optimizes the measure of effectiveness in a model. (4) Testing the validity. The solution values of the model, obtained as stated in step (3), are then tested against actual observations. In other words, effort is made to test the validity of the model used. A model is supposed to be valid if it can give a reliable prediction of the performance of the system represented through the model. If necessary, the model may be modified in the light of actual observations and the whole process is repeated till a satisfactory model is attained. The operational researcher quite often realizes that his model must be a good representation of the system and must correspond to reality which in turn requires this step of testing the validity of the model in an OR. study. In effect, performance of the model must be compared with the policy or procedure that it is meant to replace. (5) Controlling the solution. This step of an OR study establishes control over the solution by proper feedback of information on variables which might have deviated significantly. As such, the significant changes in


the system and its environment must be detected and the solution must accordingly be adjusted. This is particularly true when solutions are rules for repetitive decisions or decisions that extend over time. (6) Implementing the results. Implementing the results constitutes the last step of an OR study. Because the objective of OR is not merely to produce reports but to improve the performance of systems, the results of the research must be implemented, if they are accepted by the decision makers. It is through this step that the ultimate test and evaluation of the research is made and it is in this phase of the study the researcher has the greatest opportunity for learning.


NOTES

Thus, the procedure for an OR study generally involves some major steps viz., formulating the problem, constructing the mathematical model to represent the system under study, deriving a solution from the model, testing the model and the solution so derived, establishing controls over the solution and putting the solution to work-implementation. (Although the said phases and the steps are usually initiated in the order listed in an OR study, it should always be kept in mind that they are likely to overlap in time and to interact, i.e., each phase usually continues until the study is completed).

Flow chart showing OR approach OR approach can also be illustrated through the following flow chart: Information needs Facts, opinions and symptoms of the problem

Define the problem

Technical, financial and economic information from all sources

Determine factors affecting the problem (variables, constraints and assumptions)

Develop objective and alternative solutions Detailed information on factors

Model development

Tools of trade

Maximize or Minimize some function

Analyse alternatives and deriving optimum solution

Determination of validity. Recommended action, control and implementation

Computer service (may be used)


Operations Research

NOTES

1.5

GENERAL METHODS FOR SOLVING OR MODELS

1.5.1 Operations Research and Decision-Making Mathematical models have been constructed for OR problems and methods for solving the models are available in many cases. Such methods are usually termed as OR techniques. Some of the important OR techniques often used by decision-makers in modern times in business and industry are as follows: (i) Linear programming. This technique is used in finding a solution for optimizing a given objective such as profit maximization or cost minimization under certain constraints. This technique is primarily concerned with the optimal allocation of limited resources for optimizing a given function. The name linear programming is because of the fact that the model in such cases consists of linear equations indicating linear relationship between the different variables of the system. Linear programming technique solves product-mix and distribution problems of business and industry. It is a technique used to allocate scarce resources in an optimum manner in problems of scheduling, product-mix, and so on. Key factors under this technique include an objective function, choice among several alternatives, limits or constraints stated in symbols and variables assumed to be linear. (ii) Waiting line or queuing theory. Waiting line or queuing theory deals with mathematical study of queues. Queues are formed whenever the current demand for service exceeds the current capacity to provide that service. Waiting line technique concerns itself with the random arrival of customers at a service station where the facility is limited. Providing too much of capacity will mean idle time for servers and will lead to waste of money. On the other hand, if the queue becomes long, there will be a cost due to waiting of units in the queue. Waiting line theory, therefore, aims at minimizing the costs of both servicing and waiting. In other words, this technique is used to analyse the feasibility of adding facilities and to assess the amount and cost of waiting time. With its help we can find the optimal capacity to be installed which will lead to a sort of an economic balance between cost of service and cost of waiting. (iii) Inventory control/planning. Inventory planning aims at optimizing inventory levels. Inventory may be defined as a useful idle resource which has economic value, e.g., raw-materials, spare parts, finished products, etc. Inventory planning, in fact, answers the two questions, viz., how much to buy and when to buy? Under this technique, the main emphasis is on minimizing costs associated with holding inventories, procurement of inventories and shortage of inventories. (iv) Game theory. Game theory is used to determine the optimum strategy in a competitive situation. The simplest possible competitive situation is that of two persons playing zero-sum game, i.e., a situation in which two persons are involved and one person wins exactly what the other


loses. More complex competitive situations of the real life can also be imagined where game theory can be used to determine the optimum strategy. (v) Decision theory. Decision theory concerns with making sound decisions under conditions of certainty, risk and uncertainty. As a matter of fact, there are three different types of states under which decisions are made, viz., deterministic, stochastic and uncertainty and the decision theory explains how to select a suitable strategy to achieve some object or goal under each of these three states.


NOTES

(vi) Network analysis. Network analysis involves the determination of an optimum sequence of performing certain operations concerning some jobs in order to minimize overall time and/or cost. Programme Evaluation and Review Technique (PERT), Critical Path Method (CPM) and other network techniques such as Gantt Chart comes under Network Analysis. Key concepts under this technique are network of events and activities, resource allocation, time and cost considerations, network paths and critical paths. (vii) Simulation. Simulation is a technique of testing a model which resembles a real life situation. This technique is used to imitate an operation prior to actual performance. Two methods of simulation are there: One is Monte Carlo method of simulation and the other is System Simulation Method. The former one using random numbers is used to solve problems which involve conditions of uncertainty and where mathematical formulation is impossible, but in case of System Simulation, there is a reproduction of the operating environment and the system allows for analysing the response from the environment to alternative management actions. This method draws samples from a real population instead of drawing samples from a table of random numbers. (viii) Integrated production models. This technique aims at minimizing cost with respect to workforce, production and inventory. This technique is highly complex and is used only by big business and industrial units. This technique can be used only when sales and costs statistics for a considerable long period are available. (ix) Some other oR techniques. In addition to these, there are several other techniques such as non-linear programming, dynamic programming, search theory, and the theory of replacement. A brief mention of some of these is are as follows: (a) Non-linear programming is that form of programming in which some or all the variables are curvilinear. In other words, this means that either the objective function or constraints or both are not in the linear form. In most practical situations, we encounter nonlinear programming problems, but for computation purpose we approximate them as linear programming problems. Even then, there may remain some non-linear programming problems which may not be fully solved by presently known methods. (b) Dynamic programming refers to the systematic search for optimal solutions to problems that involve many highly complex interSelf-Instructional Material 13

Operations Research

NOTES

relations that are, moreover, sensitive to multistage effects such as successive time phases. (c) Heuristic programming, also known as discovery method, refers to step by step search toward an optimum when a problem cannot be expressed in the mathematical programming form. The search procedure examines successively a series of combinations that lead to stepwise improvements in the solution and the search stops when a near optimum has been found. (d) Integer programming is a special form of linear programming in which the solution is required in terms of integral numbers (i.e., whole numbers) only. (e) Algorithmic programming is just the opposite of heuristic programming. It may also be termed as near mathematical programming. This programming refers to a thorough and exhaustive mathematical approach to investigate all aspects of the given variables in order to obtain optimal solution. (f) Quadratic programming refers to a modification of linear programming in which the objective equations appear in quadratic form, i.e., they contain squared terms. (g) Parametric programming is the name given to linear programming when the later is modified for the purpose of inclusion of several objective equations with varying degrees of priority. The sensitivity of the solution to these variations is then studied. (h) Probabilistic programming, also known as stochastic programming, refers to linear programming that includes an evaluation of relative risks and uncertainties in various alternatives of choice for management decisions. (i) Search theory concerns itself with search problems. A search problem is characterized by the need for designing a procedure to collect information on the basis of which one or more decisions are made. This theory is useful in places in which some events are known to occur but the exact location is not known. The first search model was developed during the World War II to solve decision problems connected with air patrols and their search for submarines. Advertising agencies’ search for customers, personnel departments’ search for good executives are some examples of search theory’s application in business. (j) The theory of replacement is concerned with the prediction of replacement costs and the determination of the most economic replacement policy. There are two types of replacement models— one type of models deal with replacing equipment that deteriorate with time and the other type of models helps in establishing replacement policy for those equipment which fail completely and instantaneously.


All these techniques are not simple but involve higher mathematics. The tendency today is to combine several of these techniques and form more sophisticated and advanced programming models.

1.6

APPLICATIONS AND LIMITATIONS OF OR


NOTES

1.6.1 Applications of OR Operations research (OR) has gained increasing importance since World War II in the technology of business and industry administration. It greatly helps in tackling the intricate and complex problems of modern business and industry. OR techniques are, in fact, examples of the use of scientific method of management. The following are some applications of OR: (1) OR provides a tool for scientific analysis. OR provides the executives with a more precise description of the cause and effect relationship and the risks underlying the business operations in measurable terms and this eliminates the conventional intuitive and subjective basis on which managements used to formulate their decisions decades ago. In fact, OR replaces the intuitive and subjective approach of decision-making by an analytical and objective approach. The use of OR has transformed the conventional techniques of operational and investment problems in business and industry. As such, OR encourages and enforces disciplined thinking about organizational problems. (2) OR provides solution for various business problems. OR techniques are being used in the field of production, procurement, marketing, finance and other allied fields. Problems like how best can managers and executives allocate the available resources to various products so that in a given time the profits are maximum or the cost is minimum?, Is it possible for an industrial enterprise to arrange the time and quantity of orders of its stocks such that the overall profit with given resources is maximum? and How far is it within the competence of a business manager to determine the number of men and machines to be employed and used in such a manner that neither remains idle and at the same time the customer or the public has not to wait unduly long for service? can be solved with the help of OR techniques. Similarly, we might have a complex of industries—steel, machine tools and others—all employed in the production of one item, say, steel. At any particular time, we have a number of choices of allocating resources such as money, steel and tools for producing autos, building steel factories or tool factories. What should be the policy which optimizes the total number of autos produced over a given period? OR techniques are capable of providing an answer in such a situation. Planning decisions in business and industry are largely governed by the picture of anticipated demands. The potential long-range profits of the business may vary in accordance with different possible demand patterns. The OR techniques serve to develop a scientific basis for coping with Self-Instructional Material 15

Operations Research

NOTES

the uncertainties of future demands. Thus, in dealing with the problem of uncertainty over future sales and demands, OR can be used to generate ‘a least risk’ plan. At times there may be a problem of finding an acceptable definition of long-range company objectives. Management may be confronted with different viewpoints—some may stress the desirability of maximizing net profit, whereas others may focus attention primarily on the minimization of costs. OR techniques (especially that of mathematical programming such as linear programming) can help resolve such dilemmas by permitting systematic evaluation of the best strategies for attaining different objectives. These techniques can also be used for estimating the worth of technical innovations as also of potential profits associated with the possible changes in rules and policies. How much changes can there be in the data on which a planning formulation is based without undermining the soundness of the plan itself? How accurately must managements know cost coefficients, production performance figures and other factors before it can make planning decisions with confidence? Much of the basic data required for the development of long-range plans is uncertain. Such uncertainties though cannot be avoided, but through various OR techniques, the management can know how critical such uncertainties are and this in itself is a great help to business planners. (3) OR enables proper deployment of resources. OR renders valuable help in proper deployment of resources. For example, Programme Evaluation and Review Technique (PERT) enables us to determine the earliest and the latest times for each of the events and activities and thereby helps in the identification of the critical path. All this helps in the deployment of resources from one activity to another to enable the project completion on time. This technique, thus helps in determining the probability of completing an event or a project by a specified date. (4) OR helps in minimizing the waiting and servicing costs. The waiting line or queuing theory helps the management in minimizing the total waiting and servicing costs. This technique also analyses the feasibility of adding facilities and thereby, helps the business people to take correct and profitable decision. (5) OR enables management to decide when to buy and how much to buy. The main object of inventory planning is to achieve balance between the cost of holding stocks and the benefits from stock holding. Hence, the technique of inventory planning enables the management to decide when to buy and how much to buy. (6) OR assists in choosing an optimum strategy. Game theory is specially used to determine the optimum strategy in a competitive situation and enables the businessmen to maximize profits or minimize losses by adopting the optimum strategy. (7) OR renders great help in optimum resource allocation. Linear programming technique is used to allocate scarce resources in an optimum manner in problems of scheduling, product-mix, and so on.


This technique is popularly used by modern managements in resource allocation and in affecting optimal assignments. (8) OR facilitates the process of decision-making. Decision theory enables the businessmen to select the best course of action when information is given in a probabilistic form. Through decision tree (a network showing the logical relationship between the different parts of a complex decision and the alternative courses of action in any phase of a decision situation) technique executive’s judgement can systematically be brought into the analysis of the problems. Simulation is another important technique used to imitate an operation or a process prior to actual performance. The significance of simulation lies in the fact that it enables in finding out the effect of alternative courses of action in a situation involving uncertainty where mathematical formulation is not possible. Even complex groups of variables can be handled through this technique.


NOTES

(9) Through OR, management can know the reactions of integrated business systems. The Integrated Production Models technique is used to minimize cost with respect to workforce, production and inventory. This technique is quite complex and is usually used by companies having detailed information concerning their sales and costs statistics over a long period. Besides, various other OR techniques also help management people in taking decisions concerning various problems of business and industry. The techniques are designed to investigate how the integrated business system would react to variations in its component elements and/or external factors. (10) OR techniques help in the training/grooming of future (or would be) managers. In fact, OR techniques substitute a means for improving the knowledge and skill of youngsters in the field of management.

1.6.2 OR and Modern Business Management From what has been stated earlier, we can say that operational research renders valuable service in the field of business management. It ensures improvement in the quality of managerial decisions in all functional areas of management. The role of OR in business management can be summed up as follows: 1. OR techniques help the directing authority in optimum allocation of various limited resources viz., men, machines, money, material, time, etc., to different competing opportunities on an objective basis for achieving effectively the goal of a business unit. They help the chief of executive in broadening the management vision and perspectives in the choice of alternative strategies to the decision problems such as forecasting manpower, production capacities and capital requirements and plans for their acquisition. 2. OR is useful to the production management in (i) selecting the building site for a plant, scheduling and controlling its development and designing its layout; (ii) locating within the plant and controlling the movements of required production materials and finished goods inventories; (iii) scheduling and sequencing production by adequate preventive maintenance with optimum number of operatives by proper allocation of machines; and (iv) calculating the optimum product-mix. Self-Instructional Material 17

Operations Research

NOTES

3. OR is useful to personnel management to find out (i) the optimum manpower planning; (ii) the number of persons to be maintained on permanent or full time roll; (iii) the number of persons to be kept in a work pool intended for meeting the absenteeism; (iv) the optimum manner of sequencing and routing of personnel to a variety of jobs; and (v) in studying personnel recruiting procedures, accident rates and labour turnover. 4. OR techniques equally help marketing management to determine (i) where distribution points and warehousing should be located, their size, quantity to be stocked and the choice of customers; (ii) the optimum allocation of sales budget to direct selling and promotional expenses; (iii) the choice of different media of advertising and bidding strategies; and (iv) the consumer preferences relating to size, colour, and packaging, for various products as well as to outbid and outwit competitors. 5. OR is also very useful to financial management in (i) finding long-range capital requirements as well as how to generate these requirements; (ii) determining optimum replacement policies; (iii) working out a profit plan for the firm; (iv) developing capital-investments plans; and (v) estimating credit and investment risks. In addition to all this, OR provides business executives with an understanding of business operations which gives them new insights and capability to determine better solutions for several decision-making problems with great speed, competence and confidence. When applied on the level of management where policies are formulated, OR assists the executives in an advisory capacity but on the operational level where production, personnel, purchasing, inventory and administrative decisions are made, it provides management with a means for handling and processing information. Thus, in brief; OR can be considered as a scientific method of providing executive departments with a quantitative basis for taking decisions regarding operations under their control.

1.6.3 Limitations of OR OR—though it is a great aid to management as explained earlier—cannot be a substitute for decision-making. The choice of a criterion as to what is actually best for a business enterprise is still that of an executive who has to fall back upon his experience and judgement. This is so because of the several limitations of OR. Some important limitations are as follows: (1) The inherent limitations concerning mathematical expressions. OR involves the use of mathematical models, equations and similar other mathematical expressions. Assumptions are always incorporated in the derivation of an equation or model and such an equation or model may be correctly used for the solution of the business problems when the underlying assumptions and variables in the model are present in the concerning problem. If this caution is not given due care, then there always remains the possibility of wrong application of OR techniques. Quite often, operations researchers have been accused of having many solutions without being able to find problems that fit. (2) High costs are incurred in the use of OR techniques. OR techniques usually prove to be expensive. Services of specialized persons are 18 Self-Instructional Material

invariably called for (and along with it the use of computors) while using OR techniques. As such only big concerns can think of using such techniques. Even in big business organizations, we can expect that OR techniques will continue to be of limited use simply because they are not in many cases worth their cost. As opposed to this, a typical manager exercising intuition and judgement may be able to make a decision very inexpensively. Thus, the use of OR is a costlier affair and this constitutes an important limitation of OR.


NOTES

(3) OR does not take into consideration the intangible factors, i.e., nonmeasurable human factors. OR makes no allowance for intangible factors such as skill, attitude, vigour of the management people in taking decisions but in many instances success or failure hinges upon the consideration of such non-measurable intangible factors. There cannot be any magic formula for getting an answer to management problems; much depends upon proper managerial attitudes and policies. (4) OR is only a tool of analysis and not the complete decision-making process. It should always be kept in mind that OR alone cannot make the final decision. It is just a tool and simply suggests best alternatives, but in the final analysis many business decisions will involve human element. Thus, OR is at best a supplement rather than a substitute for management; subjective judgement is likely to remain a principal approach to decision-making. (5) Other limitations. Among other limitations of OR, the following deserve mention: (i) Bias. The operational researchers must be unbiased. An attempt to show results into a confirmation of management’s prior preferences can greatly increase the likelihood of failure. (ii) Inadequate objective functions. The use of a single objective function is often an insufficient basis for decisions. Laws, regulations, public relations, market strategies, etc., may all serve to overrule a choice arrived at in this way. (iii) Internal resistance. The implementation of an optimal decision may also confront internal obstacles such as trade unions or individual managers with strong preferences for other ways of doing the job. (iv) Competence. Competent OR analysis calls for the careful specification of alternatives, a full comprehension of the underlying mathematical relationships and a huge mass of data. Formulation of an industrial problem to an OR set programme is quite often a difficult task. (v) Reliability of the prepared solution. At times, a non-linear relationship is changed into linear for fitting the problem to the linear programming pattern. This may disturb the solution.

1.6.4 Frontiers/Scope of OR OR techniques, having their origin with war operations analysis during World War II, have since assisted defence management to a larger extent. In defence Self-Instructional Material 19

Operations Research

NOTES

management, being vitally concerned with problems of logistics such as provisioning, distributing, search and forecasting, OR has given valuable help in decision formation. Although OR originated in context of military operations, its impact nowadays can be observed in many other areas. It has successfully entered into different research areas relating to modern business and industry. In fact, it is being reckoned as an effective scientific technique to solve several managerial decisionmaking problems such as inventory management problems, resource allocation problems, problems relating to replacement of both men as well as machines, sequencing and scheduling problems, queuing problems, competitive problems, search problems and the like ones. The OR approach, though being practised by big organized business and industrial units, is equally applicable to small organizations. For example, whenever a departmental store face problems like employing additional sales girls and purchasing an additional van; we apply some techniques of operation research to minimize cost and maximize benefit for each such decision. As such it is often said that wherever there is a problem there is scope for applying operations research techniques. In recent years of organized development, operations research has successfully entered, apart from its military and business applications, into several other areas of research. For instance, the basic problem in most of the developing countries in Asia and Africa is to remove poverty as quickly as possible. There is a considerable scope to solve this problem by applying operations research techniques. Similarly, with the explosion of population and consequent shortage of food, every country is facing the problem of optimum allocation of land for various crops in accordance with the climatic conditions and the available facilities. The problem of optimal distribution of water from a source like a canal for irrigation purposes is faced by each developing country. Hence, a good amount of scientific work related to the operations research can be done in this direction. Thus, there is considerable scope of applying operations research techniques for solving problems relating to national planning. National planning can be largely improved if an optimum coordination can be arrived through central planning agency by resorting to operations research models and techniques.

Check Your Progress 4. What do you mean by formulating a problem? 5. What are the various steps involved in the procedure of an OR study? 6. Define integer programming. 7. Mention any two ways in which operations research helps personnel management. 20 Self-Instructional Material

In the field of finance, the progress of operations research has been rather slow. Financial institutions have many situations similar to those prevailing in other commercial organizations. Investment policy must maximize the return on it, keeping the factor of risk below a specified level. Long-range corporate objectives of an institution are always studied along with functional objectives of individual departments, consistency between the two being essential for corporate planning. In particular, banking institutions require precise and accurate forecasting on cash management and capital budgeting. Linear programming models for many problems and quadratic programming formulation for some complex problems may prove the usefulness of operations research techniques to credit institutions. The application of operations research techniques can as well be noticed in context of hospital management, health planning programmes, transportation system and in several other sectors. For instance, hospital management quite often faces the problem of allotting its limited resources of its multi-phased activities. Optimum allocation of resources so as to ensure a certain desirable level of service to patients can be reached through operations research. Similarly, for operating a transportation

service profitably, an optimum schedule for vehicles and crews becomes necessary. Punctuality, waiting time, total travel time, speed of the vehicles and agreement with trade unions of crews are some of the variables to be taken into account while preparing optimum schedules which can be worked out utilizing operations research techniques.


NOTES

As stated earlier, operations research is generally concerned with problems that are tactical rather than strategic in nature, i.e., its use to long-range organizational planning problems is very much limited, but one can hope that with further developments taking place in the field, ‘Operations research should be able to deal with organizations in their entirety, rather than with slices through them. Managers who currently have to treat the whole with only intuition and experience as their guides, will then have the methods of science at their disposal as well’.1 This description brings home the point that during the last four decades the scope2 of operations research has extensively been widened and hopefully shall attain new strides in the days to come. The art of systems analysis, so well developed in the military context, will spread to other contexts, notably such civil government branches as criminal justice, urban problems, housing, health care, education and social services. There is growing awareness amongst the people that unless they make themselves familiar with operations research techniques, they would not be able to understand and appreciate the problems of modern business units. With computer facilities becoming widespread, the significance and scope of operations research is likely to grow in the coming years. In fact, the future holds great promise for the growth of OR in power, scope and practical importance and utility. OR will continue its currently vigorous efforts to reach out to new arenas of exploration and application.

1.7

SUMMARY

This unit introduced you to operations research, where you have learned about its origin and development of and its characteristics. The unit described the various models and modelling techniques applied in OR. Thus, you know that a model is a simplified representation of an operation or it is a process in which only the basic aspects or the most important features of a typical problem under investigation are considered. The various steps involved in the methodology of operations research involve formulating the problem, constructing the model, deriving the solution, testing the validity, controlling the solution and finally implementing the results. The unit also described various operations research techniques, most of which are used by organizations in decision-making. Thus, to mention a few advantages, operations research techniques help in the optimum allocation of various limited resources and are also useful to the production management in many ways. Also, you should have realized that though operations research is a great help to management, it cannot be a substitute for decision-making.

1

Russell L. Ackoff and Maurice W. Sasieni, Fundamentals of Operations Research. (New York: John Wiley, 1968), 447. Self-Instructional Material 21

Operations Research

1.8 ●

Operations research: It can be considered as being the application of scientific method by interdisciplinary teams to problems involving the control of organized (man-machine) systems so as to provide solutions which best serve the purposes of the organization as a whole.

●

Operations research model: A model in OR is a simplified representation of an operation, or is a process in which only the basic aspects or the most important features of a typical problem under investigation are considered.

●

Descriptive models: They describe and predict facts and relationships among the various activities of the problem. These models do not have an objective function as part of the model to evaluate decision alternatives.

●

Normative or optimization models: They are prescriptive in nature and develop objective decision-rule for optimum solutions.

●

Iconic or physical models: They are pictorial representations of real systems and have the appearance of the real thing. An iconic model is said to be scaled down or scaled up according to the dimensions of the model which may be smaller or greater than that of the real item, e.g., city maps, houses blueprints, globe, and so on.

●

Analog models: Models in which one set of properties is used to represent another set of properties are called analog models.

●

Mathematic or symbolic models: They are most abstract in nature. They employ a set of mathematical symbols to represent the components of the real system. These variables are related together by means of mathematical equations to describe the behaviour of the system.

●

Deterministic models: They are those in which all parameters and functional relationships are assumed to be known with certainty when the decision is to be made. Linear programming and break-even models are the examples of deterministic models.

●

Probabilistic or stochastic models: These models are those in which atleast one parameter or decision variable is a random variable. These models reflect to some extent the complexity of the real world and the uncertainty surrounding it.

●

Queuing theory: Waiting line or queuing theory deals with mathematical study of queues. The queues are formed whenever the current demand for service exceeds the current capacity to provide that service.

●

Simulation: It is a technique of testing a model which resembles a real life situation. This technique is used to imitate an operation prior to actual performance.

●

Quadratic programming: This refers to a modification of linear programming in which the objective equations appear in quadratic form, i.e., they contain squared terms.

NOTES


KEY TERMS

1.9

ANSWERS TO ‘CHECK YOUR PROGRESS’

1. Operations research tries to optimize a well-defined function subject to given constraints and as such is concerned with the optimization theory.


NOTES

2. (i) It provides a logical and systematic approach to the problem. (ii) Problems under consideration become controllable through a model. 3. Models are classified in the following ways: (a) Models by function, (b) Models by structure, (c) Models by degree of abstraction, (d) Models by nature of environment, (e) Models by the extent of generality. 4. Formulating a problem consists in identifying, defining and specifying the measures of the components of a decision model. 5. The various steps involve formulating the problem, constructing the mathematical model to represent the system under study, deriving a solution from the model, testing the model and the solution so derived, establishing controls over the solution and implementation. 6. Integer programming is a special form of linear programming in which the solution is required in terms of integral numbers only. 7. OR is useful to the personnel management in finding the optimum manpower requirement and in studying personnel recruiting procedures, accident rates and labour turnover.

1.10 QUESTIONS AND EXERCISES Short-Answer Questions 1. Give two important definitions of OR. 2. Give a very brief account of the evolution of OR. 3. What are the advantages of a good OR model? 4. What are the different kinds of OR models based on structure? 5. Draw a flow chart showing the OR approach.

Long-Answer Questions 1. Define operations research and explain its main characteristics. 2. Describe in brief some of the important OR techniques used in modern business and industrial units. 3. Narrate fully the role of OR in the field of business and industry. Give examples in support of your answer. 4. Write a short note on ‘operations research’, describing some of its important techniques. 5. Discuss the limitations of OR techniques. 6. Do you think the day will come when all decisions in a business unit are made with the assistance of OR? Give reasons for your answer. Self-Instructional Material 23

Operations Research

NOTES

7. ‘Operations research is a very powerful tool and analytical process that offers the presentation of an optimum solution in spite of its limitations.’ Discuss. 8. Describe the significance and scope of OR in modern business management. 9. Discuss the various phases in solving an OR problem. 10. Describe (i) the areas of application of OR, (ii) the state of OR in India.

1.11 FURTHER READING Bierman, Harold, Charles Bonini and W.H. Hausman. 1973. Quantitative Analysis for Business Decisions. Homewood (Illinois): Richard D. Irwin. Gillett, Billy E. 2007. Introduction to Operations Research. New Delhi: Tata McGraw-Hill. Lindsay, Franklin A. New Techniques for Management Decision Making. New York: McGraw-Hill. Herbert, G. Hicks. 1967. The Management of Organisations. New York: McGrawHill. Massie, Joseph L. 1986. Essentials of Management. New Jersey: Prentice-Hall. Ackoff, R.L. and M.W. Sasieni. 1968. Fundamentals of Operations Research. New York: John Wiley & Sons Inc. Enrick, Norbert Lloyd. 1965. Management Operations Research. New York: Holt, Rinehart & Winston. Kothari, C.R. 1982. An Introduction to Operational Research. New Delhi: Vikas Publishing House. Kalavathy, S. 2002. Operations Research. New Delhi: Vikas Publishing House.


Linear Programming

UNIT 2 LINEAR PROGRAMMING Structure 2.0 Introduction 2.1 Unit Objectives 2.2 Linear Programming Formulation

NOTES

2.2.1 General Formulation of LPP 2.2.2 Matrix Form of LPP

2.3 Graphical Method 2.3.1 2.3.2 2.3.3 2.3.4 2.3.5

Procedure for Solving LPP by Graphical Method General Formulation of LPP Matrix Form of LPP Some Important Definitions Canonical or Standard Forms of LPP

2.4 Big M Method 2.5 Simplex Method 2.5.1 Introduction 2.5.2 Definition

2.6 Duality in LP: Conversion of Primal to Dual 2.6.1 Formulation of Dual Problem 2.6.2 Definition of Dual Problem 2.6.3 Important Results in Duality

2.7 2.8 2.9 2.10 2.11

2.0


INTRODUCTION

Linear programming deals with the optimization (maximization or minimization) of a function of variables known as objective functions. It is subject to a set of linear equalities and/or inequalities known as constraints. Linear programming is a mathematical technique which involves the allocation of limited resources in an optimal manner on the basis of a given criterion of optimality. In this unit, the properties of Linear Programming Problems (LPP) are discussed. The graphical method of solving a LPP is applicable where two variables are involved. The most widely used method for solving LP problems consisting of any number of variables is called the simplex method developed by G. Dantzig in 1947 and made generally available in 1951.

2.1

UNIT OBJECTIVES

After going through this unit, you will be able to: ● Describe the procedure for mathematical formulation of LPP ● Explain the procedure for solving LPP by graphical method ● Describe the steps involved in solving LPP using the Big M method Self-Instructional Material 25

Operations Research

● ●

NOTES

Understand the procedure of solving LPP using the simplex method Understand the concept of duality in LP

2.2

LINEAR PROGRAMMING FORMULATION

The procedure for mathematical formulation of an LP problem consists of the following steps: Step 1 Write down the decision variables of the problem. Step 2 Formulate the objective function to be optimized (Maximized or Minimized) as a linear function of the decision variables. Step 3 Formulate the other conditions of the problem such as resource limitation, market constraints and interrelations between variables as linear inequations or equations in terms of the decision variables. Step 4 Add the non-negativity constraint from the considerations so that the negative values of the decision variables do not have any valid physical interpretation. The objective function, the set of constraint and the non-negative constraint together form a linear programming problem.

2.2.1 General Formulation of LPP The general formulation of the LPP can be stated as follows: In order to find the values of n decision variables x1x2 ... xn maximize or minimize the objective function. Z

c1 x1

c2 x2 … cn xn

a11 x1 a12 x2 a21 x1 a22 x2 : ai1 x1 ai 2 x2 : am1 x1 am 2 x2

...(1)

a1n xn ( a2 n xn (

)b1 )b2

ain xn (

)bi

amn xn (

...(2)

)bn

where constraints may be in the form of inequality ≤ or ≥ or even in the form an equation (=) and finally satisfy the non negative restrictions x1

0, x2

0… xn

...(3)

0

2.2.2 Matrix Form of LPP The LPP can be expressed in the matrix form as follows: Maximize or Minimize Z = Cx → Objective function subject to Ax (≤=≥) b Constant equation b > 0, x ≥ 0 Non-negativity restrictions, where 26 Self-Instructional Material

x = ( x1 x2

xn )

c = (c1 c2 " cn )

Linear Programming

b=

b1 b2

A

bm

a11a12 a21a22 : am1am 2

a1n a2 n amn

NOTES

Example 2.1: A manufacturer produces two types of models M1 and M2. Each model of the type M1 requires 4 hours of grinding and 2 hours of polishing; whereas each model of the type M2 requires 2 hours of grinding and 5 hours of polishing. The manufacturers have 2 grinders and 3 polishers. Each grinder works 40 hours a week and each polisher works for 60 hours a week. Profit on M1 model is Rs 3.00 and on model M2 is Rs 4.00. Whatever is produced in a week is sold in the market. How should the manufacturer allocate his production capacity to the two types of models, so that he may make the maximum profit in a week? Solution: Decision variables: Let X1 and X2 be the number of units of M1 and M2 model. Objective function: Since the profit on both the models are given, we have to maximize the profit, viz.

Max Z = 3X1 + 4X2 Constraints: There are two constraints, one for grinding and the other for polishing.

The number of hours available on each grinder for one week is 40 hrs. There are two grinders. Hence, the manufacturer does not have more than 2 × 40 = 80 hours of grinding. M1 requires 4 hours of grinding and M2 requires 2 hours of grinding. The grinding constraint is given by 4 x1

2 x2

80

Since there are 3 polishers, the available time for polishing in a week is given by 3 × 60 = 180. M1 requires 2 hours of polishing and M2 requires 5 hours of polishing. Hence, we have 2x1 + 5x2 ≤ 180. Finally we have Max Z = 3x1 + 4x2 Subject to 4 x1 2 x2 2 x1 5 x2 x1 x2

80

180

0

Example 2.2: A company manufactures two products A and B. These products are processed in the same machine. It takes 10 minutes to process one unit of product A and 2 minutes for each unit of product B and the machine operates for a maximum of 35 hours in a week. Product A requires 1 kg. and B 0.5 kg. of raw material per unit the supply of which is 600 kg. per week. Market constraint on product B is known to be 800 units every week. Product A costs Rs 5 per unit and sold at Rs 10. Product B costs Rs 6 per unit and can be sold in the market at a unit price of Rs 8. Determine the number of units of A and B per week to maximize the profit. Self-Instructional Material 27

Operations Research

Solution: Decision variables: Let x1 and x2 be the number of products A and B.

NOTES

Objective function: Costs of product A per unit is Rs 5 and sold at Rs 10 per unit. ∴ Profit on one unit of product A

= 10 – 5 = 5 ∴ x1 units of product A contributes a profit of Rs 5 profit contribution from one unit of product

B=8–6=2 :. x2 units of product B contribute a profit of Rs 2 ∴ The objective function is given by

Max Z

5 x1

2 x2

Constraints: Time requirement constraint is given by 10 x1

2 x2

(35 60)

10 x1

2 x2

21.00

Raw material constraint is given by x1

0.5 x2

600

Market demand on product B is 800 units every week. ∴ x2 ≥ 800

The complete LPP is Max Z

5 x1

2 x2

Subject to 10 x1 2 x2 x1 0.5 x2 P x2

2100 600

800

x1 , x2

0

Example 2.3: A person requires 10,12, and 12 units of chemicals A, B respectively for his garden. A liquid product contains 5,2 and 1 units of A, B and C respectively per jar. A dry product contains 1,2 and 4 units of A, B, C per carton. If the liquid product sells for Rs 3 per jar and the dry product sells for Rs 2 per carton, how many of each should be purchased, in order to minimize the cost and meet the requirements? Solution: Decision variables: Let X1 and X2 be the number of units of liquid and dry products. Objective function: Since the cost for the products are given we have to minimize the cost.

Min Z = 3x1 + 2x2 28 Self-Instructional Material

Constraints: As there are three chemicals and its requirement are given, we have three constraints for these three chemicals. 5 x1

x2

10

2 x1 2 x2

12

x1 4 x2

12

Linear Programming

NOTES

Finally, the complete LPP is Min Z = 3x1 + 2x2 Subject to 5 x1 x2 10

2 x1 2 x2

12

x1 4 x2

12

x1 , x2

0

Example 2.4: A paper mill produces two grades of paper, namely X and Y. Because of raw material restrictions, it cannot produce more than 400 tonnes of grade X and 300 tonnes of grade Y in a week. There are 160 production hours in a week. It requires 0.2 and 0.4 hours to produce a tonne of products X and Y respectively with corresponding profits of Rs 200 and Rs 500 per tonne. Formulate this as a LPP to maximize profit and find the optimum product mix. Solution: Decision variables: Let x1 and x2 be the number of units of two grades of paper X and Y. Objective function: Since the profit for the two grades of paper X and Y are given, the objective function is to maximize the profit.

Max Z = 200x1 + 500x2 Constraints: There are 2 constraints one w.r.t. to raw material, and the other w.r to concerning production hours.

Max Z = 200x1 + 500x2 Subject to x1

400

x2

300

0.2 x1 0.4 x2

160

Non-negative restriction x1 x2 ≥ 0 Example 2.5: A company manufactures two products A and B. Each unit of B takes twice as long to produce as one unit of A and if the company were to produce only A it would have time to produce 2000 units per day. The availability of the raw material is sufficient to produce 1500 units per day of both A and B combined. Product B requiring a special ingredient only 600 units can be made per day. If A fetches a profit of Rs 2 per unit and B a profit of Rs 4 per unit, find the optimum product mix by graphical method. Solution: Let x1 and x2 be the number of units of the products A and B respectively.

The profit after selling these two products is given by the objective function Max Z = 2x1 + 4x2


Operations Research

NOTES

Since the company can produce at the most 2000 units of the product in a day and type B requires twice as much time as that of type A, production restriction is given by x1

2 x2

2000

Since the raw materials are sufficient to produce 1500 units per day both A and B combined we have x1 x2 1500. There are special ingredients for the product B; we have x2 ≤ 600. Also, since the company cannot produce negative quantities x1 ≥ 0 and x2 ≥ 0. Hence, the problem can be finally put in the form: Find x1 and x2 such that the profits Z = 2x1 + 4x2 is maximum. x1 2 x2 P

2000

x2 1500

x1 Subject to

x2 x1 , x2

600 0

Example 2.6: A firm manufacturers 3 products A, B and C. The profits are Rs 3, Rs 2 and Rs 4 respectively. The firm has 2 machines and the followign is the required processing time in minutes for each machine on each product.

Machines C D

Product A B C 4 3 5 3 2 4

Machine C and D have 2000 and 2500 machine minutes respectively. The firm must manufacture 100 units of A, 200 units of B and 50 units of C but no more than 150 A’s. Set up an LP problem to maximize the profit. Solution: Let x1, x2, x3 be the number of units of the product A, B, C respectively.

Since the profits are Rs 3, Rs 2 and Rs 4 respectively, the total profit gained by the firm after selling these three products is given by Z

3 x1

2 x2

4 x3

The total number of minutes required in producing these three products at machine C is given by 4x1 + 3x2 + 5x3 and at machine D is given by 2x1 + 2x2 + 4x3. The restrictions on the machine C and D are given by 2000 minutes and 2500 minutes. 4 x1 3 x2

5 x3

2000

2 x1

4 x3

2500

2 x2

Also, since the firm manufactures 100 units of A, 200 units of B and 50 units of C but not more than 150 unit of A the further restriction becomes 30 Self-Instructional Material

Linear Programming

100 ≤ x1 ≤ 150 200 ≤ x2 ≥ 0 50 ≤ x3 ≥ 0

NOTES

Hence, the allocation problem of the firm can be finally put in the form: x1, x2, x3 so as to maximize Z = 3x1 + 2x2 + 4x3 Subject to the constraints F

i n d

t h e

v

a l u

e

o f

4 x1 3 x2

5 x3 2000

2 x1 2 x2

4 x3 2500

100

x1 150, 200

x2

0,50

x3

0

Example 2.7: A farmer has a 100 acre farm. He can sell all tomatoes, lettuce or radishes and can raise the price to obtain Re 1.00 per kg. for tomatoes, Rs 0.75 a head for lettuce and Rs 2.00 per kg. for radishes. The average yield per acre is 2000 kg. of tomatoes, 3000 heads of lettuce and 1000 kgs of radishes. Fertilizers are available at Rs.0.50 per kg. and the amount required per acre is 100 kg each for tomatoes and lettuce and 50 kgs for radishes. Labour required for sowing, cultivating and harvesting per acre is 5 man-days for tomatoes and radishes and 6 man-days for lettuce. A total of 400 man-days of labour are available at Rs 20,000 per man-day. Formulate this problem as a linear programming model to maximize the farmer’s total profit. Solution: Let x1, x2, x3 be the area of his farm to grow tomatoes, lettuce and radishes respectively. The farmer produces 2000x1 kgs of tomatoes, 3000x2 heads of lettuce and 1000x3 kg. of radishes. ∴ The total sales of the farmer will be = Rs (2000x1 + 0.75 × 3000x2 + 2 × 1000x3) ∴ Fertilizer expenditure will be = Rs 20 (5x1 + 6x2 + 5x3) :. Farmer’s profit will be

Z = Sale (in Rs) – total expenditure (in Rs). = (2000x1 + 0.75 × 3000x2 + 2 × 1000x3) –0.5 × [100(x1 + x2) + 50x2] –20 × (5x1 + 6x2+5x3) Z = 1850x1 + 2080x2 + 1875x3 Since the total area of the farm is restricted to 100 acres x1 + x2 + x3 ≤ 100. Also, the total man-days labour is restricted to 400 man-days 5 x1 6 x2

5 x3

400

Hence, the farmer’s allocation problem can be finally put in the form: Find the value of x1, x2 and x3 so as to maximize Z = 1850x1 + 2080x2 + 1875x3 Self-Instructional Material 31

Operations Research

Subject to x1

x2

5 x1 6 x2

NOTES

x3

100

5 x3

400

x1 x2 x3

0

Example 2.8: An electric appliance company produces two products: refrigerators and ranges. Production takes place in two separate departments I and II. Refrigerators are produced in department I and ranges in department II. The company’s two products are sold on a weekly basis. The weekly production can not exceed 25 refrigerators and 35 ranges. The company regularly employs a total of 60 workers in two departments. A refrigerator requires 2 man-weeks labour while a range requires 1man-week labour. A refrigerator contributes a profit of Rs 60 and a range contributes a profit of Rs 40. How many units of refrigerators and ranges should the company produce to realize the maximum profit? Formulate this as a LPP. Solution: Let x1 and x2 be the number of units of refrigerators and the ranges to be produced.

Each refrigerator and range contributes a profit of Rs 60 and Rs 40 ∴ The objective function is to maximize Z = 60x1 + 40x2. There are 2 constraints which are imposed on weekly production and labours. Since the weekly production cannot exceed 25 refrigerators and 35 ranges, x1

25

x1

35

A refrigerator requires 2 man-weeks of labour and a range requires 1 man-week of labour and the total number of workers is 60. 2x1 + x2 ≤ 60. Non-negativity restrictions. Since the number of refrigerators and ranges produced cannot be negative, we have x1 ≥ 0 and x2 ≥ 0. Hence, the production of refrigerator and ranges problem can be finally put in the form: Find the value of x1 and x2 so as to maximize Z = 60x1 + 4x2 Subject to x1 25 2 x1

x2

35

x2

60

and x1 , x2

2.3

0

GRAPHICAL METHOD

A simple linear programming problem with two decision variables can be easily solved by the graphical method. 32 Self-Instructional Material

2.3.1 Procedure for Solving LPP by Graphical Method

Linear Programming

The steps involved in the graphical method are as follows. Step 1

Consider each inequality constraint as an equation.

Step 2 Plot each equation on the graph as each will geometrically represent a straight line.

NOTES

Step 3 Mark the region. If the inequality constraint corresponding to that line is O then the region below the line lying in the first quadrant (due to non-negativity of variables) is shaded. For the inequality constraint P sign, the region above the line in the first quadrant is shaded. The points lying in common region will satisfy all the constraints simultaneously. The common region, thus obtained, is called the feasible region. Step 4

Assign an arbitrary value, say zero, for the objective function.

Step 5 Draw a straight line to represent the objective function with the arbitrary value (i.e., a straight line through the origin). Step 6 Stretch the objective function line till the extreme points of the feasible region. In the maximization case, this line will stop farthest from the origin, passing through at least one corner of the feasible region. In the minimization case, this line will stop nearest to the origin and passes through at least one corner of the feasible region. Step 7 Find the coordinates of the extreme points selected in step 6 and find the maximum or minimum value of Z. Note: As the optimal values occur at the corner points of the feasible region, it is enough to calculate the value of the objective function of the corner points of the feasible region and select the one which gives the optimal solution, i.e., in the case of maximization problem the optimal point corresponds to the corner point at which the objective function has a maximum value and in the case of minimization, the corner point which gives the objective function the minimum value is the optimal solution. Example 2.9: Solve the following LPP by graphical method.

Minimize Z = 20X1 + 10X2 Subject ot X1 + 2X2 ≤ 40 3X1 + X2 ≥ 30 4X1 + 3X2 ≥ 60 X1, X2 ≥ 0 Solution:

Replace all the inequalities of the constraints by equation

X1 + 2X2 = 40

If X1 = 0 ⇒ X2 = 20

If X2 = 0 ⇒ X1 = 40 :. X1 + 2X2 = 40

passes through (0, 20) (40, 0)

3X1 + X2 = 30


4X1+ 3X2 = 60


Plot each equation on the graph. Self-Instructional Material 33

Operations Research

NOTES

The feasible region is ABCD. C and D are the points of intersection of lines X1+ 2X2 = 40, 3X1+ X2 = 30 and 4X1+ 3X2= 60, X1 + X2 = 30 On solving we get C = (4, 18) D=(6, 12) Corner points

Value of Z =20X1 + 10X2

A (15, 0)

300

B (40, 0)

800

C(4, 18)

260

D (6, 12)

240 (Minimum value)

∴ The minimum value of Z occurs at D (6, 12). Hence, the optimal solution is X1 = 6, X2= 12. Example 2.10: Find the maximum value of Z = 5X1 + 7X2.

Subject to the constraints, X1 + X2 ≤ 4 3X1 + 8X2 ≤ 24 10X1 + 7X2 ≤ 35 X1, X2 > 0 Solution: Replace all the inequalities of the constraints by forming equations.

X1 + X2 = 4


3X1 + 8X2 = 24 passes through (0, 3) (8, 0) 10X1 + 7X2 = 35 passes through (0, 5) (3.5, 0) Plot these lines in the graph and mark the region below the line since the inequality of the constraint is ≤ and is also lying in the first quadrant.


Linear Programming

NOTES

The feasible region is OABCD. B and C are points of intersection of lines. X1 + X2 = 4, 10X1 + 7X2 = 35 and 3X1 + 8X2 = 24, X1 + X2 = 4 On solving we get, B = (1.6, 2.3) C = (1.6, 2.4) Corner points

Value of Z = 5X1 + 7X2

O (0, 0)

0

A (3.5, 0)

17.5

B (1.6, 2.3)

25.1

C (1.6, 2.4)

24.8 (Maximum value)

D (0, 3) 21 ∴ The maximum value of Z occurs at C (1.6, 2.4) and the optimal solution is X1 = 1.6, X2 = 2.4. Example 2.11: A company produces 2 types of hats. Every hat of type A requires twice as much labour time than the second type B. If the company proclues only hat B then it can produce a total of 500 hats a day. The market limits daily sales of the hat A and hat B to 150 and 250 hats. The profits on hat A and B are Rs 8 and Rs 5 respectively. Solve graphically to get the optimal solution. Solution: Let X1 and X2 be the number of units of type A and type B hats respectively.

Max Z = 8X1 + 5X2 Subject to

2X1 + 2X2 ≤ 500

X1 ≥ 150 X2 ≥ 250 X1,X2 ≥ 0. First rewrite the inequality of the constraint into an equation and plot the lines in the graph. Self-Instructional Material 35

Operations Research

2X1 + X2 = 500


X1 = 150

passes through (150, 0)

X2= 250

NOTES

passes through (0, 250)

We mark the region below the lines lying in the first quadrant since the inequality of the constraints is ≤ . The feasible region is OABCD and B and C are points of intersection of lines 2X1 + X2 = 500,

X1 = 150 and

2X1 + X2 = 500,

X2 = 250.

On solving, we get B = (150,200) C = (125,250)

Corner points O (0,0)

Value of Z = 8X1 + 5X2 0

A (150,0)

1200

B (150,200)

2200

C (125,250)

2250 (Maximum Z = 2250)

D (0,250)

1250

The maximum value of Z is attained at C (125, 250). ∴ The optimal solution is X1 = 125, X2 = 250.

i.e., the company should produce 125 hats of type A and 250 hats of type B in order to get the maximum profit of Rs 2250. Example 2.12: By graphical method, solve the following LPP.


5X1 + 4X2 ≤ 200 3X1 + 5X2 ≤ 150 5X1 + 4X2 ≥ 100 8X1 + 4X2 ≥ 80

and 36 Self-Instructional Material

X1, X2 ≥ 0

Solution:

Linear Programming

NOTES

Feasible region is given by OABCD. Corner points

Value of [Z 3X1 ] 4X2

O (20, 0)

60

A (40, 0)

120

B (30.8, 11.5)

138.4

C (0, 30)

120

D (0, 25)

100

(Maximum value)

∴ The maximum value of Z is attained at B (30.8,11.5). ∴ The optimal solution is X1 = 30.8, X2 = 11.5. Example 2.13: Use graphical method to solve the LPP.

Maximize Z = 6X1 + 4X2 Subject to

–2X1 + X2 ≤ 2

X1 – X2 ≤ 2 3X1 + 2X2 ≤ 9 X1, X2 ≥ 0 Solution:


Operations Research

Feasible region is given by ABC. Corner points

NOTES


A (2, 0)

12

B (3,0)

18

C (13/5, 3/5)

98/5 = 19.6 (Maximum value)

The maximum value of Z is attained at c (13/5, 3/5). ∴ The optimal solution is X1= 13/5, X2= 3/5. Example 2.14: Use graphical method to solve the LPP.


5X1 + X2 ≥ 10 X1 + X2 ≥ 6 X1 + 4X2 ≥12 X1, X2 ≥ 0.

Solution:

Corner points


A (0, 10)

20

B (1,5)

13

C (4, 2)

16

D (12, 0)

36

(Minimum value)

Since the minimum value is attained at B (1,5) the optimum solution is X1, X2 = 5. Note: In this problem if the objective function is maximization then the solution is unbounded, as the maximum value of Z occurs at infinity.

Some more cases There are some linear programming problems which may have (i) A unique optimal solution (ii) An infinite number of optimal solution 38 Self-Instructional Material

(iii) An unbounded solution

Linear Programming

(iv) No solution The following examples will illustrate these cases. Example 2.15: Solve the LPP by graphical method.

NOTES


5X1 + 2X2 ≤ 1000

3X1 + 2X2 ≤ 900 X1 + 2X2 ≤ 500 and X1 + X2 ≥ 0 Solution:

The solution space is given by the feasible region OABC. Value of Z = 100X1+ 40X2

Corner points O (0, 0)

0

A (200, 0)

20,000

B (125, 187.5)

20,000 (Max value of Z)

C (0, 250)

10,000

∴ The maximum value of Z occurs at two vertices A and B.

Since there are infinite number of points on the line joining A and B which gives the same maximum value of Z. Thus, there are infinite number of optimal solutions for the LPP. Example 2.16:

Solve the following LPP.


X1 + X2 ≥ 1 X1 + X2 ≥ 3 X1, X2 ≥ 0

Solution: The solution space is unbounded. The value of the objective function at the vertices A and B are Z (A) = 6, Z (B) = 6. But, there exist points in the convex region for which the value of the objective function is more than 8. In Self-Instructional Material 39

Operations Research

fact, the maximum value of Z occurs at infinity. Hence, the problem has an unbounded Solution:

NOTES

No feasible solution When there is no feasible region formed by the constraints in conjunction with non-negativity conditions, then no solution to the LPP exists. Example 2.17:

Solve the following LPP.

Max Z = X1 + X2 Subject to the constraints X1+X2 ≤ 1 –3X1 + X2 ≥ 3 X1, X2 ≥ 0

Solution:

There being no point (X1, X2) common to both the shaded regions, we cannot find a feasible region for this problem. So the problem cannot be solved. Hence, the problem has no solution.

2.3.2 General Formulation of LPP The general formulation of the LPP can be stated as follows: Maximize or Minimize Z – C1X1 + C2X2 + ... + CnXn 40 Self-Instructional Material

...(1)

Subject to m constraints

Linear Programming

⎧ a11 X 1 + a12 X 2 + ⎪ ⎪ a21 X 1 + a22 X 2 + ⎪ ⎪⎪ ⎨ ai1 X 1 + ai 2 X 2 + ⎪ ⎪ ⎪am1 X 1 + am 2 X 2 + ⎪ ⎪⎩

+ a1 j X j +

+ am X n (≤ = ≥)b1 ⎫ ⎪ + a2 j X j + + a2 n X n (≤ = ≥)b2 ⎪ ⎪ ⎪⎪ + a1 j X j + + a1n X n (≤ = ≥)bi ⎬ ⎪ ⎪ + amj X j + + amn X n (≤ = ≥)bm ⎪ ⎪ ⎪⎭

NOTES

... (2)

In order to find the values of n decision variables X1 X2... Xn to maximize or minimize the objective function and the non-negativity restrictions X1 ≥ 0, X2 ≥ 0 ≥ Xn ≥ 0

...(3)

2.3.3 Matrix Form of LPP The linear programming problem can be expressed in the matrix form as follows: Maximize or Minimize Z = CX ⎛≤⎞ Subject to ⎜⎜ = ⎟⎟ b ⎜≥⎟ ⎝ ⎠ X ≥ 0. ⎛ x1 ⎞ ⎜ ⎟ ⎜ x2 ⎟ Where x = ⎜ ⎟ , b = ⎜⎜ x ⎟⎟ ⎝ n⎠ ⎛ a11 ⎜ a21 and A = ⎜ ⎜ ⎜⎜ a ⎝ m1

a12 a22 am 2

⎛ b1 ⎞ ⎜ ⎟ ⎜ b2 ⎟ ⎜ ⎟ , C = (C1C2 – Cn) ⎜⎜ b ⎟⎟ ⎝ m⎠ a1n ⎞ ⎟ a2 n ⎟ ⎟ ⎟ amn ⎟⎠

2.3.4 Some Important Definitions 1. A set of values X1 , X2 ... Xn which satisfies the constraints (2) of the LPP is called its solution. 2. Any solution to a LPP which satisfies the non-negativity restrictions (3) of the LPP is called its feasible solution. 3. Any feasible solution which optimizes (Minimizes or Maximizes) the objective function (1) of the LPP is called its optimum solution. Self-Instructional Material 41

Operations Research

NOTES

4. Given a system of m linear equations with n variables (m < n) any solution which is obtained by solving for m variables keeping the remaining n-m variables zero is called a basic solution. Such m variables are called basic variables and the remaining variables are called non-basic variables. 5. A basic feasible solution is a basic solution which also satisfies (3), that is all basic variables are non-negative. Basic feasible solutions are of two types: (a) Non-degenerate: A non-degenerate basic feasible solution is the basic feasible solution which has exactly m positive Xi (i = 1, 2 ......m) ie None of the basic variables are zero. (b) Degenerate: A basic feasible solution is said to degenerate if one or more basic variables are zero. 6. If the value of the objective function Z can be increased or decreased indefinitely such solutions are called unbounded solutions.

2.3.5 Canonical or Standard Forms of LPP The general LPP can be put in the following forms, namely canonical and standard forms. In the standard form, irrespective of the objective function, namely maximize or minimize, all the constraints are expressed as equations. Moreover, RHS of each constraint and all variables are non-negative.

Characteristics of the Standard Form (i) The objective function is of maximization type. (ii) All constraints are expressed as equations. (iii) The right hand side of each constraint is non-negative. (iv) All variables are non-negative. In the canonical form, if the objective function is of maximization, all the constraints other than non-negativity conditions are ‘≤’ type. If the objective function is of minimization, then all the constraints other than non-negative condition are ‘≥’ type.

Characteristics of the Canonical Form (i) The objective function is of maximization type. (ii) All constraints are of (≤) type. (iii) All variables Xi are non-negative. Note: Check Your Progress 1. Define linear programming. 2. State the general formulation of LPP. 3. How do you state an LPP in the matrix form?

(i) Minimization of a function Z is equivalent to maximization of the negative expression of this function, i.e., Min Z = –Max (–Z). (ii) An inequality in one direction can be converted into an inequality in the opposite direction by multiplying both sides by (-1). (iii) Suppose we have the constraint equation a11 x1+a12x2 +...... +a1n xn = b1


This equation can be replaced by two weak inequalities in opposite directions

Linear Programming

a11 x1+ a12 x2 +...... + am xn ≤ b1 a11 x1+ a12 x2 +...... + a1n xn ≥ b1 (iv) If a variable is unrestricted in sign, then it can be expressed as a difference of two non-negative variables, i.e., if X1 is unrestricted in sign, then X1 = X1′ – X1′′, where X1′ , X1′ – X1′′ are ≥ 0.

NOTES

(v) In the standard form, all the constraints are expressed in equation, which is possible by introducing some additional variables called slack variables and surplus variables so that a system of simultaneous linear equations is obtained. The necessary transformation will be made to ensure that bi ≥ 0.

Definition (i) If the constraints of a general LPP be n

∑a j =1

ij

xi ≤ bi (i = 1, 2...m)

Then the non-negative variables Si which are introduced to convert the inequalities n

(≤) to the equalities

∑a j =1

ij

xi + Si = bi (i = 1, 2...m) are called slack variables.

Slack variables are also defined as the non-negative variables which are added in the LHS of the constraint to convert the inequality ‘≤’ into an equation. (ii) If the constraints of a general LPP be n

∑a j =1

ij

x j ≥ bi (i = 1, 2...m)

Then, the non-negative variables S i which are introduced to convert the inequalities ≥ to the equalities

n

∑a j =1

ij

x j – Si = bi (i = 1, 2...m) are called surplus

variables. Surplus variables are defined as the non-negative variables which are removed from the LHS of the constraint to convert the inequality (≥) into an equation.

2.4

BIG M METHOD

The following steps are involved in solving an LPP using the Big M method. Step 1

Express the problem in the standard form.

Step 2 Add non-negative artificial variables to the left side of each of the equations corresponding to constraints of the type ≥ or =. However, addition of these artificial variable causes violation of the corresponding constraints. Therefore, we would like to get rid of these variables and would not allow them to appear in the final solution. This is achieved by assigning a very large penalty (–M for maximization and M for minimization) in the objective function. Self-Instructional Material 43

Operations Research

NOTES

Step 3 Solve the modified LPP by the simplex method, until any one of the three cases may arise.

1. If no artificial variable appears in the basis and the optimality conditions are satisfied, then the current solution is an optimal basic feasible solution. 2. If at least one artificial variable in the basis at zero level and the optimality condition is satisfied then the current solution is an optimal basic feasible solution (though degenerated solution). 3. If at least one artificial variable appears in the basis at positive level and the optimality condition is satisfied, then the original problem has no feasible solution. The solution satisfies the constraints but does not optimize the objective function, since it contains a very large penalty M and is called pseudo-optimal solution. Note: While applying the simplex method, whenever an artificial variable happens to leave the basis, we drop that artificial variable and omit all the entries corresponding to its column from the simplex table. Example 2.18: Use penalty method to

Maximize Z = 3x1 + 2x2 Subject to the constraints 2x1 + x2 ≤ 2 3x1 + 4x2 ≥ 12 x1, x2 ≥ 0 Solution: By introducing slack variable S1 ≥ 0, surplus variable S2 ≥ 0 and artificial variable A1 ≥ 0, the given LPP can be reformulated as:

Maximize Z = 3x1, + 2x2 + 0S1 + 0S2 – MA1, Subject to 2x1 + x2 + S1 = 2 3x1 + 4x2 – S2 + A1 = 12 The starting feasible solution is S1= 2, A1= 12. Initial table Cj

3

2

0

0

–M

CB

B

xB

x1

x2

S1

S2

A1

Min xy/x2

0

S1

2

2

1

1

0

0

2/1 = 2

–M

A1

12

3

4

0

–1

1

12/4 = 3

Zj

–12M

–3M

–4M

0

M

–M

Zj – Cj

–

0

M

0

–3M – 3 –4M – 2

Since some of the Zj – Cj ≤ 0, the current feasible solution is not optimum. Choose the most negative Zj – Cj = –4M–2. ∴ x2 variable enters the basis, and the basic variable S1 leaves the basis.


Cj

3

2

0

0

–M

CB

B

xB

x1

x2

S1

S2

A1

2

x1

2

2

1

1

0

0

–M

A1

4

–5

0

–4

–1

1

4 – 4M 4 + 5M

2

2+4M

M

–M

5M + 1

0

4M+2

M

0

Zj Zj – Cj

Linear Programming

NOTES

Since all Zj – Cj ≥ 0 and an artificial variable appears in the basis at positive level, the given LPP does not possess any feasible solution. But the LPP possesses a pseudo-optimal solution. Example 2.19: Solve the LPP

Minimize Z = 4x1 + x2 Subject to 3x1 + x2 = 3 4x1 + 3x2 ≥ 6 x1 + 2x2 ≤ 3 x1, x2 ≥ 0 Solution: Since the objective function is minimization, we covert it into maximization using

Min Z = –Max (–z) Maximize Z = –4x1 – x2 Subject to 3x1 + x2 = 3 4xl + 3x2 ≥ 6 x1 +2x2 ≤ 3 x1, x2 ≥ 0 Convert the given LPP into the standard form by adding artificial variables A1, A2, surplus variable S1 and slack variable S2 to get the initial basic feasible solution. Maximize Z = –4x1 – x2 + 0S1 + 0S2 – MA1 – MA2 Subject to 3x1 + x2 + A1 = 3 4x1 + 3x2 – S1 + A2 = 6 x1 + 2x2 + S2 = 3 The starting feasible solution is A1 = 3, A2 = 6, S2 = 3.


Operations Research

NOTES

Initial solution Cj

–4

–1

–M

0

–M

0

xB x2

CB

B

xB

x1

x2

A1

S1

A2

S2

Min

–M

A1

3

3

1

1

0

0

0

3/3 = 1

–M

A2

6

4

3

0

–1

1

0

6/4 = 1.5

←0

S2

3

1

2

0

0

0

1

3/1 = 3

Zj Zj – Cj

–9M

–7M –7M + 4

–4M –4M + 1↑

–M 0

M M

–M 0

0 0

Since some of the Zj – Cj ≥ 0, the current feasible solution is not optimum. As Z1 – C1 is most negative, x1 enters the basis and the basic variable A2 leaves the basis. First iteration Cj

–4

–1

–M

0

–M

0

xB

x1

x2

A1

S1

A2

S2

x Min B

CB

B

–M

A1

3/2

5/2

0

1

0

0

–1/2

3/5

← –M

A2

3/2

5/2

0

0

–1

1

–3/2

3/5

–1

x2

3/2

3/2

1

0

0

0

1.2

3

Zj

–3M–3/2

–5M–1/2

–1

–M

+M

–M

2M–1/2

–5M–7/2↑

0

0

M

0

2M–1/2

Zj – Cj

x1

Since Z1 – C1 is negative, the current feasible solution is not optimum. Therefore, x1 variable enters the basis and the artificial variable A2 leaves the basis.

Second iteration Cj

–4

–1

–M

0

0

xB

x1

x2

A1

S1

S2

x Min B

1

1

0

CB

B

–M

A1

0

0

0

1

← –M

x1

3 5

1

0

0

2 5

–1

x2

6 5

0

1

0

1 5

–4

–1

–M

Zj Zj – Cj

18 5

0


0

0

3 5 4 5

–M+

9 5

–M

9 ↑ 5

x1

–

–

Since Z4 – C4 is most negative, S1 enters the basis and the artificial variable A1 leaves the basis.

Linear Programming

Third iteration NOTES Cj

–4

–1

0

0

xB

x1

x2

S1

S2

x Min B

1

0

CB

B

←0

S1

0

0

0

1

–4

x1

3 5

1

0

0

–1

x2

6/5

0

1

0

1

–4

–1

0

–1/5

0

0

0

–1/5↑

Zj

18 5

Zj – Cj

x2

1 5

– 6/5

Since Z4 – C1 is most negative, S2 enters the basis and S1 leaves the basis.

Fourth iteration Cj

–4

–1

0

0

CB

B

xB

x1

x2

S1

S2

0

S2

0

0

0

1

1

–4

x1

3/5

1

0

–1

x2

6/5

0

1

1 5 –1

Zj

–18/5

–4

–1

1/5

0

0

0

1/5

0

Zj – Cj

0 0

Since all Zj – Cj ≥ 0, the solution is optimum and is given by x1= 3/5, x2= 6/5, and Max Z = –18/5. ∴ Min Z = –Max (–Z) = 18/5. Example 2.20:

Solve by Big M method.

Maximize Z = x1 + 2x2 + 3x3 – x4 Subject to

x1 + 2x2 + 3x3 = 15

2x1 + x2 + 5x3 = 20 x1 + 2x2 + x3 + x4 =10 Solution: Since the constraints are equations, introduce artificial variables A1, A2 ≥ 0. The reformulated problem is given as follows.

Maximize Z = x1 + 2x2 + 3x3 – x4 – MA1 – MA2 Subject to

x1 + 2x2 + 3x3 + A1 = 15


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2x, + x2 + 5x3 + A2 = 20 x1 +2x2 + x3 + x4 =10 Initial solution is given by A1 = 15, A2 = 20 and x4 = 10.

NOTES

Initial table Cj

1

2

3

–1

–M

–M xB x3

CB

B

xB

x1

x2

x3

x4

A1

A2

Min

–M

A1

15

1

2

3

0

1

0

15/3= 5

← –M

A2

20

2

1

5

0

0

1

20/5 = 4

–1

x4

10

1

2

1

1

0

0

10/1 = 10

Zj

–35M

–3M

–3M

–3M–1

–1

–M

–M

–10

–1

–2

–3M–2

–3M–4

–3M–4↑

0

0

0

Zj – Cj

Since Z3 – C3 is most negative, x3 enters the basis and the basic variable A2 leaves the basis.

First iteration Cj

1

2

3

–1

–M xB x2

CB

B

xB

x1

x2

x3

x4

A1

Min

← –M

A1

3

–1/5

7/5

0

0

1

3 15 7/5 7

3

A2

4

2/5

1/5

1

0

0

4 1/ 5

20 1

–1

x4

6

3/5

9/5

0

1

0

6 9/5

30 9

Zj

–3M+6

6 5

3

–1

–M

16 ↑ 5

0

0

0

Z j – Cj

1 M 5

3 5

1 M 5

3 5

7 M 5 7 M 5

Since 02 – C2 is most negative x2 enters the basis and the basic variable A1 leaves the basis.


Linear Programming

Second iteration CB

B

Cj

1

2

3

–1

xB

x1

x2

x3

x4

Min

xB x1

2

x2

15/7

1 7

1

0

0

–

3

x3

25/7

3/7

0

1

0

25/3

← –1

x4

15/7

6 7

0

0

1

15/6

Zj

90 7

1 –6/7 ↑ 7

2

3

–1

0

0

0

Zj – Cj

NOTES

Since Z1 – C1 = –6/7 is negative, the current feasible solution is not optimum. Therefore, x1 enters the basis and the basic variable x4 leaves the basis. Cj

1

2

3

–1

CB

B

xB

x1

x2

x3

x4

2

x2

15/6

0

1

0

1/6

3

x3

15/6

0

0

1

3/6

1

x4

15/6

1

0

0

7/6

Zj

15

1

3

3

0

0

Zj – Cj

1 X

0X

B 1

Since all Zj – Cj ≥ 0, the solution is optimum and is given by x1 = x2 = x3 = 15/6 = 5/2, and Max Z = 15.

2.5

SIMPLEX METHOD

2.5.1 Introduction Simplex method is an iterative procedure for solving LPP in a finite number of steps. This method provides an algorithm which consists of moving from one vertex of the region of feasible solution to another in such a manner that the value of the objective function at the succeeding vertex is less or more as the case may be than at the previous vertex. This procedure is repeated and since the number of vertices is finite, the method leads to an optimal vertex in a finite number of steps or indicates the existence of unbounded solution.

2.5.2 Definition (i) Let XB be a basic feasible solution to the LPP. Max Z = CX Subject to AX = b Self-Instructional Material 49

Operations Research

and X ≥ 0, such that it satisfies XB = B–1b where B is the basis matrix formed by the column of basic variables.

NOTES

The vector CB = (CB1, CB2 … CBm) where CBj are components of C associated with the basic variables is called the cost vector associated with the basic feasible solution XB. (ii) Let XB be a basic feasible solution to the LPP. Max Z = CX where AX = b and X ≥ 0. Let CB be the cost vector corresponding to XB. For each column vector aj in A1, which is not a column vector of B, let m

aj

aij b j i 1 m

Then the number Z j

CBi aij i 1

is called the evaluation corresponding to aj and the number (Zj – Cj) is called the net evaluation corresponding to j.

Simplex algorithm For the solution of any LPP by simplex algorithm the existence of an initial basic feasible solution is always assumed. The steps for the computation of an optimum solution are as follows: Step 1 Check whether the objective function of the given LPP is to be maximized or minimized. If it is to be minimized then we convert it into a problem of maximization by

Min Z = –Max (–Z) Step 2 Check whether all bi (i = 1,2 … m) are positive. If any one of bi is negative then multiply the inequation of the constraint by –1 so as to get all bi to be positive. Step 3 Express the problem in standard form by introducing slack/surplus variables, to convert the inequality constraints into equations. Step 4 Obtain an initial basic feasible solution to the problem in the form XB = B–1b and put it in the first column of the simplex table. Form the initial simplex table as follows


Step 5 Compute the net evaluations Zj – Cj by using the relation

Linear Programming

Zj – Cj = CB (aj – cj). Examine the sign of Zj – Cj . (i) If all Zj – Cj ≥ 0, then the initial basic feasible solution XB is an optimum basic feasible solution.

NOTES

(ii) If at least one Zj – Cj > 0, then proceed to the next step as the solution is not optimal. Step 6 (To find the entering variable i.e., key column)

If there are more than one negative Zj – Cj, choose the most negative of them. Let it be Zr – Cr for some j = r. This gives the entering variable Xr and is indicated by an arrow at the bottom of the rth column. If there are more than one variables having the same most negative Zj – Cj then any one of the variable can be selected arbitrarily as the entering variable. (i) If all Xir ≤ 0 (i = 1, 2 … m) then there is an unbounded solution to the given problem. (ii) If at least one Xir > 0 (i = 1, 2 … m) then the corresponding vector Xr enters the basis. Step 7 (To find the leaving variable or key row)

Compute the ratio (XBi /Xir, Xir>0). If the minimum of these ratios be XBi /Xkr, then choose the variable Xk to leave the basis called the key row and the element at the intersection of key row and key column is called the key element. Step 8 Form a new basis by droping the leaving variable and introducing the entering variable along with the associated value under CB column. Convert the leading element to unity by dividing the key equation by the key element and all other elements in its column to zero by using the formula

New Element = Old Element –

Product of elements in key row and key column key element

Step 9 Go to step (5) and repeat the procedure until either an optimum solution is obtained or there is an indication of unbounded solution. Example 2.22: Use simplex method to solve the LPP.

Max Z = 3X1 + 2X2 Subject to X 1 X1

X2

4

X2

2

X1, X 2

0

Solution: By introducing the slack variables S1, S2, convert the problem into the standard form.

Max Z = 3X1 + 2X2 + 0S1 + 0S2 Self-Instructional Material 51

Operations Research

Subject to

NOTES X1 1 1

X1

X2

S1

4

X1

X2

S2

2

X 1 , X 2 , S1 , S2

0

X 2 S1 1 1 1 0

S2 0 1

X1 X2 S1 S2

4 2

An initial basic feasible solution is given by XB = B–1b, where B = I2, XB = (S1 S2). i.e., (S1 S2) = I2 (4,2) = (4,2)

Initial simplex table Zj = CB aj ⎛ 0⎞ Z1 − c1 = CB a1 − c1 = ⎜ ⎟ ⎝ 0⎠ ⎛ 0⎞ Z 2 − c2 = C B a2 − c2 = ⎜ ⎟ ⎝ 0⎠

⎛ 1⎞ ⎜ ⎟ − 3 = −3 ⎝ 1⎠ (1 − 1) − 2 = −2

⎛ 0⎞ Z 3 − c3 = CB a3 − c3 = ⎜ ⎟ (1 0) − 0 = −0 ⎝ 0⎠ ⎛ 0⎞ Z 4 − c4 = CB a4 − c4 = ⎜ ⎟ (0 1) − 0 = −0 ⎝ 0⎠

Cj

3

2

0

0 XB X1

CB

B

XB

X1

X2

S1

S2

0

S1

4

1

1

1

0

4/1 = 4

←0

S2

2

1

–1

0

1

2/1 = 2

Zj

0

0

0

0

0

–3↑

–2

0

0

Zj – Cj

Min

Since there are some Zj – Cj=0, the current basic feasible solution is not optimum. Since Z1 – C1= –3 is the most negative, the corresponding non-basic variable X1 enters the basis.


The column corresponding to this X1 is called the key column. To find the ratio = Min

= Min

X Bi , X ir X ir

Linear Programming

0

NOTES

4 2 , = 2 which corresponds to S2. 2 1

∴ The leaving variable is the basic variable S2. This row is called the key row. Convert the leading element X21 to units and all other elements in its column, i.e., (X1) to zero by using the formula: New element = Old element – Product of elements in key row and key column key element

To apply this formula, first we find the ratio, namely the element to be zero key element

1 1 1

Apply this ratio, for the number of elements that are converted in the key row. Multiply this ratio by key row element shown as follows. 1×2 1×1 1 × –1 1×0 1×1 Now, subtract this element from the old element. The elements to be converted into zero, is called the old element row. Finally we have 4–1×2=2 1–1×1=0 1 – 1 × –1 = 2 1–1×0=1 0 – 1 × 1 = –1

∴ The improved basic feasible solution is given in the following simplex table


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NOTES

First iteration Cj

3

2

0

0

0 XB X1

CB

B

xB

x1

x2

S1

S2

S3

Min

0

S1

12

4

3

1

0

0

12/4 = 4

0

S2

8

4

1

0

1

0

8/4 = 2

←0

S3

8

4

–1

0

0

1

8/4 = 2

Zj

0

0

0

0

0

0

–3↑

–2

0

0

0

Zj – Cj

Since Z2 – C2 is most negative, X2 enters the basis. To find Min

Min

XB , X i2 X i2

2 , 2

0

=1.

This gives the outgoing variables. Convert the leading element into 1. This is done by dividing all the elements in the key row by 2. The remaining element is multiplied by zero using the formula as shown below. –½ is the common ratio.Put this ratio 5 times and multiply each ratio by key row element. 1 2 2 1 0 2 1 2 2

–1/2 × 2 –1/2 × –1 Subtract this from the old element. All the row elements which are converted into zero are called the old elements. 2

1 2 2

3

1 – (–1/2 × 0) = 1 –1 – (–1/2 × 2) = 1 0 – (–1/2 × 1) = 1/2 1 – (–1/2 × 1) = 1/2 54 Self-Instructional Material

Linear Programming

Second iteration Cj

1

1

3

0

0

CB

B

XB

X1

X2

S3

S1

S2

0

S1

2

2

3/2

0

1

–1/2

3

X3

1

1

1/2

1

0

1/2

Zj

3

3

3/2

3

0

3/2

2

1/2

0

0

3/2

Zj – Cj

1 1

,

NOTES

Since all Zj – Cj ≥ 0, the solution is optimum. The optimal solution is Max Z = X1 = 3, and X2 = 1. Example 2.23: Solve the LPP

Max Z = 3X1 + 2X2 Subject to 4 X 1 3 X 2 12 4 X1 X 2 8 4 X1

X2

8

X1, X 2

0

Solution: Convert the inequality of the constriant into an equation by adding slack variables S1, S2, S3.

Max Z = 3X1 + 2X2 + 0S1 + 0S2 + 0S3 4 X1 3 X 2

S1

12

4 X1

X2

S2

8

4 X1

X2

S3

8

X 1 , X 2 , S1 , S 2 , S3

0

Subject to

X1 4 4 4

X 2 S1 3 1 1 0 1 0

S2 0 1 0

S3 0 0 1

X1 X2 S1 S2 S3

12 8 8


Operations Research

NOTES

Initial table Cj

3

2

0

0

0 XB X1

CB

B

XB

X1

X2

S1

S2

S3

Min

0

S1

12

4

3

1

0

0

12/4 = 3

0

S2

8

4

1

0

1

0

8/4 = 2

←0

S3

8

4

–1

0

0

1

8/4 = 2

Zj

0

0

0

0

0

0

–3↑

–2

0

0

0

Zj – Cj

⎛ xB ⎞ ∴ Z1 – C1 is most negative, X1 enters the basis. And the min ⎜ , xil > 0 ⎟ = min ⎝ xil ⎠ (3, 2, 2) = 2 gives S3 as the leaving variable.

Convert the leading element into 1, by dividing key row element by 4 and the remaining elements into 0.

First iteration Cj

3

2

0

0

0 XB X2

B

XB

X1

X2

S1

S2

S3

0

S1

4

0

4

1

0

–1

4/4 = 1

←0

S2

0

0

2

0

1

–1

0/2 = 1

3

X1

2

1

–1/4

0

0

1/4

–

Zj

6

3

–3/4

0

0

3/4

0

–11/4↑

0

0

3/4

Zj – Cj


Min

CB

8

4 8 4

0

12

4

4 4 4

0

4

4 4 4

1

4 4

3

4 4

0

4 0 4

1

4 0 1 4

1 2

0

4 8 4

4

0

1 4

4 1 0 1 4 0

4 1 4

Since Z 2

4 0 0 4 1

C2

0 3

4

4 1 4

Linear Programming

0

NOTES

1

is the most negative, x2 enters the basis.

To find the outgoing variable, find min

xB

xi 2 , xi 2

0

⎛4 0 ⎞ min ⎜ , , − 1⎟ = 0 ⎝4 2 ⎠

First iteration

Therefore S2 leaves the basis. Convert the leading element into 1 by dividing the key row elements by 2 and remaining element in that column into zero using the formula. New element = old element – Product of elements in key row and key column key element C1

3

2

0

0

0 X3

C3

B

X3

X1

X2

S1

S2

S1

Min S 3

←0

S1

4

0

0

1

–2

1 ○

4/1 = 1

2

X2

0

0

1

0

½

–½

–

3

X1

2

1

0

0

1/8

1/8

2/1/8 = 16

Zj

6

3

2

0

11/8

–5/8

0

0

0

11/8

–5/8↑

Zj – Cj Second iteration

Since Z5 – C5 = –5/8 is most negative, S3 enters the basis and Min

XB , S13 S13

Min

4 2 , 1 1 1/18

4.

Therefore, S1 leaves the basis. Convert the leading element into one and the remaining elements into zero.


Operations Research

Third iteration Cj

3

2

0

0

0

CB

B

XB

X1

X2

S1

S2

S3

0

S1

4

0

0

1

–2

1

2

S2

2

0

1

1/2

–1/2

0

3

X1

3/2

1

0

–1/x

3/8

0

Zj

17/2

3

2

5/8

1/8

0

0

0

5/8

1/8

0

NOTES

Zj – Cj

Since all Zj – Cj ≥ 0, the solution is optimum and it is given by X1 = 3/2, X2 = 2 and Max Z = 17/2. Example 2.24: Using simplex method solve the LPP.

Maximize Z = X1 + X2 + 3X3 Subject to 3 X 1 2 X 2 X 3 2 X1 X 2 2 X 3

2

X1 , X 2 , X 3

0

3

Solution: Rewrite the inequality of the constraints into an equation by adding slack variables.

Max Z = X1 + X2 + 3X3 + 0S1 +0S2 Subject to 3 X 1 2 X 2 X 3 S1 2 X 1 X 2 2 X 3 S2

3 2

Initial basic feasible solution is X1

X2

S1

3, S2

X1 3 2 1

X2 2 1 1

X3

2 and Z

X3 1 2 3

S1 1 0 0

0

S2 0 1 0 Cj

3

2

0

0

0

0 Min

XB X3

CB

B

XB

X1

X2

X3

S1

S2

0

S1

3

3

2

1

1

0

3/1 = 3

←0

S2

2

2

1

2

0

1

2/2 = 1

Zj

0

0

0

0

0

0

–1

–1

–3

0

0

Zj – Cj 58 Self-Instructional Material

0

Since Z3 – C3 = –3 is the most negative, the variable X3 enters the basis. The column corresponding to X3 is called the key column. (To determine the key row or the leaving variable, find min 3 1

3,

2 2

1

xB , xi 3 xi 3

0 Min

Linear Programming

NOTES

1 ).

Therefore, the leaving variable is the basic variable S2, the row is called the key row and the intersection element 2 is called the key element. Convert this element into 1 by dividing each element in the key row by 2 and the remaining elements in that key column into zero using the formula New element = old element – Product of elements in key row and key column key element

First iteration Cj

1

1

3

0

0

CB

B

XB

X1

X2

X3

S1

S2

0

S1

2

2

3/2

0

1

–1/2

3

X3

1

1

1/2

1

0

1/2

Zj

3

3

3/2

3

0

3/2

2

1/2

0

0

3/2

Zj – Cj

Since all Zj – Cj ≥ 0, the solution is optimum and it is given by X1 = 0, X2 = 0, X3 = 1, Max Z = 3.

2.6

DUALITY IN LP: CONVERSION OF PRIMAL TO DUAL

Every LPP (called the primal) is associated with another LPP (called its dual). Either of the problem can be considered as primal with the other as dual. The importance of the duality concept is due to two main reasons: (i) If the primal contains a large number of constraints and a smaller number of variables, the labour of computation can be considerably reduced by converting it in to the dual problem and then solving it. (ii) The interpretation of the dual variables from the cost or economic point of view, proves extremely useful in making future decisions in the activities being programmed.


Operations Research

2.6.1 Formulation of Dual Problem For formulating a dual problem, first we bring the problem in the canonical form. The following changes are used in formulating the dual problem.

NOTES

(1) Change the objective function of maximization in the primal into that of minimization in the dual and vice versa. (2) The number of variables in the primal will be the number of constraints in the dual and vice versa. (3) The cost coefficients C1 , C2 ... Cn in the objective function of the primal will be the RHS constant of the constraints in the dual and vice versa. (4) In forming the constraints for the dual, we consider the transpose of the body matrix of the primal problem. (5) The variables in both problems are non-negative. (6) If the variable in the primal is unrestricted in sign, then the corresponding constraint in the dual will be an equation and vice versa.

2.6.2 Definition of Dual Problem Let the primal problem be Max Z = C1x1 + C2x2 + ... + Cnxn Subject to

a11 x1 a12 x2 … a1n xn

b1

a21 x1 a22 x2 … a2 n xn

b2

am1 x1 am 2 x2 … amn xn

bm

x1 , x2

xn

0

Dual: The dual problem is defined as M

i n

Z1 = b1w1 + b2w2 + ... + bmwm

Subject to a11w1 a21w2 … am1wm

C1

a12 w1 a22 w2 … am 2 wm

C2

a1n w1 a2 n w2 … amn wn

Cn

w1 , w2

wm

0

where w1, w2, w3 ... wm are called dual variables. Example 2.25: Write the dual of the following primal LP problem.

Max Z = x1+ 2x2 + x3


Linear Programming

Subject to

2 x1

x2

2 x1

x3

x2

4 x1

2

5 x3

x2

6

x3

6

x1 , x2 , x3

0

NOTES

Solution: Since the problem is not in the canonical form, we interchange the inequality of the second constraint.

Max Z = x1+ 2x2 + x3 2 x1

Subject to

x3

2

5 x3

6

x3

6

and x1 , x2 , x3

0

2 x1

x2 x2

4 x1

x2

Dual: Let w1, w2, w3 be the dual variables.

Min Z1 = 2w1+ 6w2 + 6w3 Subject to 2 w1

2 w2

4 w3

1

w2

w3

2

w1 5w2

w3

1

w1 , w2 , w3

0

w1

Example 2.26: Find the dual of the following LPP.

Max Z = 3x1– x2 + x3 Subject to

4 x1

x2

8

x2

3x3

12

5 x1 6 x3

13

x1 , x2 , x3

0

8 x1

Solution: Since the problem is not in the canonical form, we interchange the inequality of the second constraint.

Max Z = 3x1– x2 + x3 Subject to

4 x1

x2

8 x1

x2 3 x3

5 x1 0 x2

0 x3 6 x3

x1 , x2 , x3

8 12 13 0

Max Z = Cx Subject to Ax B x 0 Self-Instructional Material 61

Operations Research

NOTES

⎛ x1 ⎞ ⎛ 8 ⎞ ⎜ ⎟ ⎜ ⎟ C = (3 − 11) × ⎜ x2 ⎟ b = ⎜ −12 ⎟ ⎜x ⎟ ⎜ 13 ⎟ ⎝ 3⎠ ⎝ ⎠ ⎛ 4 −1 0 ⎞ ⎜ ⎟ A = ⎜ −8 −1 −3 ⎟ ⎜ ⎟ ⎝ 5 0 −6 ⎠ Dual: Let w1, w2, w3 be the dual variables. The dual problem is

Min Z1 = bTW Subject to ATW ≥ CT and W ≥ 0 ⎛ w1 ⎞ ⎜ ⎟ i.e. Min Z1 = (8 −12 13) ⎜ w2 ⎟ ⎜w ⎟ ⎝ 3⎠ ⎛ 4 −8 5 ⎞ ⎛ w1 ⎞ ⎛ 3 ⎞ Subject to ⎜ −1 −1 0 ⎟ ⎜ w2 ⎟ ≥ ⎜ −1⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ 0 −3 6 ⎟ ⎜ w ⎟ ⎜ 1 ⎟ ⎝ ⎠⎝ 3⎠ ⎝ ⎠

Min Z1 = 8w1 – 12w2 + 13w3 Subject to 4 w1 − 8w2 + 5w3 ≥ 3 − w1 − w2 + 0w3 ≥ −1 0 w1 − 3w2 + 6 w3 ≥ 1 w1 , w2 , w3 ≥ 0

Example 2.27: Write the dual of the following LPP

Min Z = 2x2 + 5x3 x1

Subject to

x2

2

2 x1

x2

6 x3

6

x1

x2

3x3

4

x1 , x2 , x3

0

Solution: Since the given primal problem is not in the canonical form, we interchange the inequality of the constraint. Also, the third constraint is an equation. This equation can be converted into two inequations.

Min Z = 0x1 + 2x2 + 5x3


Linear Programming

Subject to

x1

x2

0 x3

2

2 x1

x2

6 x3

6

x1

x2

3 x3

4

x1

x2

3 x3

4

x1 , x2 , x3

0

NOTES

Again on rearranging the constraint, we have M

i n

Z = 0x1 + 2x2 + 5x3

Subject to

x1

x2

0 x3

2 x1

x2

6 x3

x1

x2

3 x3

x1

x2

3 x3

2 6 4 4

x1 , x2 , x3

0

Dual: Since there are four constraints in the primal, we have four dual variables, namely w1, w2, w3′, w3′′.

Max Z′ = 2w1 – 6w2 + 4w3′ – 4w3′′ Subject to

w1 2 w2

w3

w3

0

w3

w3

2

3w3 3w3

5

w1 , w2 , w3 , w3

0

w1

w2

0 w1 6 w2

Let w3

w3

Max Z

w3

2 w1 6 w2

Subject to w1 2 w2 w1 w2

4( w3

w3 )

( w3

w3 ) 0

( w3

w3 ) 2

Finally, we have, 0 w1 6 w2 Max Z Subject to

2 w1 6 w2

3( w3

5

4 w3

w1 2w2

w3

0

w1 w2

w3

2

3w3

5

0 w1 6w2

w3 )

w1 , w2

0, w3 is unrestricted.

Example 2.28: Give the dual of the following problem

Max Z = x + 2y


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Subject to 2 x 3 y 3x 4 y

4 5

x

0 and y unrestricted.

NOTES Solution: Since the variable y is unrestricted, it can be expressed as y y,y

y

y ,

0 . On reformulating the given problem, we have

Max Z

x 2( y

Subject to

y )

2x 3 ( y

y )

3x 4 ( y

y )

5

3x 4 ( y

y )

5

x, y , y

4

0

Since the problem is not in the canonical form we rearrange the constraints. Max Z = x + 2y′ – 2y′′ Subject to

2x 3 y 3x 4 y

3y 4y

3x 4 y

4y

4 5 5.

Dual: Since there are three variables and three constraints, in dual we have three variables, namely w1, w2′, w2′′.

Min Z Subject to

4 w1 5w2

5w2

2w1 3w2 3w2

1

3w1 4 w2

4 w2

2

3w1 4 w2

4 w2

w1 , w2 , w2

2 0

Let w2 = w2′ – w2′′, so that the dual variable w2 is unrestricted in sign. Finally the dual is Min Z 1

Subject to

4 w1 5 ( w2

5w2 )

2 w1 3 ( w2

w2 ) 1

3w1

4 ( w2

w ) 2

3w1

4 ( w2

w )

i.e., Min Z 1 = –4w1 + 5w2 Subject to

2w1 3w2

1

3w1

2

4w2

3w1 4w2 64 Self-Instructional Material

2

2

w1 ≥ 0 and w2 is unrestricted.

Linear Programming

i.e., Min Z 1 = –4w1 + 5w2 Subject to

2w1 3w2

1

3w1 4w2

2

3w1 4w2

2

NOTES

i.e., Min Z 1 = –4w1 + 5w2 Subject to –2w1 + 3w2 ≥ 1 –3w1 + 4w2 = 2, w1 ≥ 0 and w2 is unrestricted. Example 2.29: Write the dual of the following primal LPP.

Min Z = 4x1 + 5x2 – 3x3 Subject to

x3

22

3 x1 5 x2

2 x3

65

x1 7 x2

4 x3

120

x1

x2

x1 + x2 ≥ 0 and x3 is unrestricted. Solution: Since the variable x3 is unrestricted, x3

x3

x3 . Also, bring the

problem into the canonical form by rearranging the constraints. Min Z = 4x1 + 5x2 – 3 ( x3 x1

Subject to

x3 )

( x3

x2 x1

x2

x3

x3 )

22

x3

22

3 x1 5 x2

2 ( x3

x3 )

65

x1 7 x2

4 ( x3

x3 )

120

x1 , x2 , x3

0

x3

Min Z = 4x1 + 5x2 – 3 x3 3 x3 Subject to

x1

x2

x3

x3

22

x1

x2

x3

x3

22

3 x1 5 x2

2 x3

2 x3

65

x1 7 x2

4 x3

4 x3

120

x1 , x2 , x3

x3

0

Dual: Since there are four constraints in the primal problem, in dual there are

four variables, namely w1 , w2 , w2 , w3 so that the dual is given by Max Z

22( w1

w1 )

65w2 120 w3 Self-Instructional Material 65

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w1 w1 3w2

w3

4

w1 w1 5w2

7 w3

5

w1 w1 2 w2

4w3

w1

w1 2 w2

4w3

3

w1

w1 2 w2

4w3

3

w1 , w1 , w2 , w3

0

Subject to

NOTES

Let w1

w1

i.e. Max Z

3

w11 , i.e., the variable w1 is unrestricted. 22( w1

Subject to

w1 )

w1

65w2 120 w3

w3

4

w1 5w2

7 w3

5

( w1 w1 ) 2w2

4w3

3

( w1 w1 ) 2w2

4w3

3

w1

i.e.,

w1 3w2

Max Z w1 3w2

22 w1

w3

4

w1 5w2

7 w3

5

w1 2w2

4 w3

3

w1 2w2

4 w3

3

Subject to

65w2 120 w3

Finally we have Min Z Subject to

22 w1

65w2 120 w3

w1 3w2

4w3

4

w1 5w2

7 w2

5

w1 2w2

4 w3

3

w2, w3 ≥ 0 and w1 is unrestricted.

2.6.3 Important Results in Duality 1. The dual of the dual is primal. 2. If one is a maximization problem then the other is a minimization one. 3. The necessary and sufficient condition for any LPP and its dual to have an optimal solution is that both must have feasible solution.


4. Fundamental duality theorem states if either the primal or dual problem has a finite optimal solution, then the other problem also has a finite optimal solution and also the optimal values of the objective function in both the problems are the same ie Max Z = Min Z. The solution of the other problem can be read from the Zj – Cj row below the columns of

Linear Programming

slack, surplus variables. 5. Existence Theorem states that, if either problem has an unbounded solution then the other problem has no feasible solution. 6. Complementary slackness theorem: According to which

NOTES

(i) If a primal variable is positive, then the corresponding dual constraint is an equation at the optimum and vice versa. (ii) If a primal constraint is a strict inequality then the corresponding dual variable is zero at the optimum and vice versa. Example 2.30: Find the maximum of Z = 6x + 8y

Subject to 5 x 2 y 20 x 2 y 10 x, y 0 by solving its dual problem. Solution: The dual of the given primal problem is given below. As there are two constraints in the primal, we have two dual variables, namley w1 and w2.

Min Z′ = 20wl + 10w2 Subject to

5w1

w2

6

2w2

8

w1 , w2

0

2w1

We solve the dual problem using the big M method. Since this method involves artificial variables, the problem is reformulated and we have Max Z

20 w1 10 w2

Subject to

5w1

0 S1 0 S 2

MA1

w2

S1

A1

6

2 w2

S2

A2

8

w1 , w2 , S1 , S 2 , A1 , A2

0

2 w1

MA2

Cj

1

1

3

0

0

0

0 Min

xB w1

CB

B

XB

w1

w2

S1

S2

A1

A2

←–M

A1

6

5

1

–1

0

1

0

6/5 = 1.02

–M

A2

8

2

2

0

–1

0

1

8/2 = 4

Zj

–14M

–7M

–3M

M

M

–M

–M

–7M+20

–3M–10

M

M

0

0

1/5

–1/5

0

–

0

Zj – Cj

↑ –20

w1

6/5

1

6/5×5/1 = 6


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NOTES

Cj

–20

–10

0

0

–M

–M

CB

B

XB

w1

–w2

S1

S2

A1

A2

Min

xB w1

←–M

A2

28 5

0

8/5

2/5

–1

–

1

28 5

5 8

=

Zj

–

25 M–24 8

Z j – Cj

–20

0

–4–8/5M

–

8 M+6 5

4–2/5M

M

–

–M

2 M 5

M

–

0

4–

25 8

↑ –20

w1

1 2

1

0

–1/4

1 8

–

–

–10

w2

7 2

0

1

1/4

–5 8

–

–

Zj

–45

–20

–10

5/2

15/4

–

–

0

0

5/2

15/4

Z j – Cj

Since all Zj – Cj ≥ 0, the solution is optimum. Therefore, the optimal solution of dual is w1 = 1/2, w2 = 7/2, min Z′ = –45 The optimum solution of the primal problem is given by the value of Zj – Cj in the optimal table corresponding to the column surplus variables S1 and S2.

∴x=

5 15 y= 2 4

5 15 Max Z = 6 × + 8 × = 45 2 4

Example 2.31: Apply the principle of duality to solve the LPP.

Max Z = 3x1 + 2x2 subject to x1 + x2 ≥ 1 x1 + x2 ≤ 10 x2 ≤ 3, x1, x2 ≥ 0 Solution: First we convert the given (primal) problem into its dual. As there are four constraints in the primal problem we have four variables w1 w2 w3 w4 in its dual. We convert the given problem into its canonical form by rearranging some of the constraints.

Max Z = 3x2 + 2x2 Subject to –x1 – x2 ≤ –1 x1 + x2 ≤ 7 68 Self-Instructional Material

x1 + 2x2 ≤ 10

0xl + x2 ≤ 3

Linear Programming

x1: x2 ≥ 0 Dual

Min

NOTES Z' = –w1 + 7w2 + 10w3 + 3w4 –w1 + w2 + w3 + 0w4 ≥ 3

Subject to

–w1 + w2 + 2w3 + w4 ≥ 2 w1, w2, w3, w4 ≥ 0 We apply Big M method to get the solution of the dual problem, as it involves artificial variables. Max Subject to

Z1 = w1 – 7w2 – 10w3 – 3w4 + 0S1 + 0S2 – MA1 – MA2 –w1 + w2 + w3 + 0w1 – S1 + A1 = 3 –w1 + w2 + 2w3 + w4 – S2 + A2 =2 w1, w2, w3, w4, S1 S2 A1, A2 ≥ 0

Since all Zj – Cj ≥ 0 (refer Table 2.1) the solution is optimum. The optimal solution of the dual problem is w1 = w3 = w4 = 0, w2 = 3, Min Z1 = –21. Also from the optimum simplex table of the dual problem, the optimal solution of the primal problem is given by the value Zj – Cj corresponding to the column of surplus variables S1 and S2

∴

x1 = 7, x2 = 0. Max Z = 21.

Example 2.32: Write down the dual of the following LPP and solve it. Hence or

otherwise write down the solution of the primal. Max

Z = 4x1 + 2x2

Subject to x1 + x2 ≥ 3 x1 – x2 ≥ 2 x1, x2 ≥ 0 Solution The dual of the given (primal) problem is as follows. First we convert the given problem into its canonical form by rearranging the constraints.

Max Subject to

Z' = 4x1 +2x2 –x1 – x2 ≤ –3 –x1 + x2 ≤ –2 Self-Instructional Material 69

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Table 2.1

NOTES


x1, x2 ≥ 0

Linear Programming

Dual Min

Z' = –3w1 – 2w2 –w1 – w2 ≥ 4

Subject to

NOTES

–w1 + w2 ≥ 2 w1, w2 ≥ 0 Introducing the surplus variables S1, S2 > 0 and artificial variables A1 A2 ≥ 0 the problem becomes Max

Z' = 3w1 + 2w2 + 0S1 + 0S2 – MA1 – MA2

Subject to

–w1 – w2 – S1 + A1 = 4 –w1 + w2 – S2 + A2 = 2 w1, w2, Sl, S2, Al, A2 ≥ 0

∴ all Zj – Cj ≥ 0 and a artificial variable A1 is in the basis at positive level. Thus, the dual problem does not possess any optimum basic feasible solution. Consequently, there exists no finite optimum solution to the given LP problem or the solution of the given LPP is unbounded. Example 2.33: Prove using duality theory that the following LPP is feasible but has no optimal solution. M

i n

Z = x1 – x2 + X3

Subject to x1 – x3 ≥ 4 x1 – x2 + 2x3 ≥ 4 and x1, x2, x3 ≥ 0 Solution:

Given primal LPP is

Min

Z = x1 – x2 + x3

Subject to

x1 + 0x3 – x3 ≥ 4 x1 – x2 + 2x3 ≥ 3 xl, x2, x3 ≥ 0

Check Your Progress 4. What are unbounded solutions? 5. Define slack variables. 6. State the complementary slackness theorem. Self-Instructional Material 71

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Dual Since there are two constraints, there are two variables w1, w2 in the dual, given by

Max

NOTES

Subject to

Z’ = 4w] + 3w2 w1 + w2 ≤ 1 0w1 – w2 ≤ –1 –w1 + 2w2 ≤ 1 w1, w2 ≥ 1

To solve the dual problem Max Subject to

Z′ = 4w1 + 3w2 w1 + w2 + S1 = 1 0w1 + w2 – S2 + A1 = 1 –w1 + 2w2 + S3 = 1

where S1, S3 are the slack variables, S2 the surplus variable and A1 the artificial variable.

Since all Zj – Cj ≥ 0 and an artificial variable appears in the basis at positive level, the dual problem has no optimal basic feasible solution.

∴ There exists no finite optimum solution to the given primal LPP (Unbounded solution).


2.7

SUMMARY

This unit has introduced you to linear programming, which is a decision-making technique under given constraints on the assumption that the relationships among the variables representing different phenomena happen to be linear. You have learned that any solution to a LPP which satisfies the non-negativity restrictions of the LPP is called its feasible solution. You have also learned about the simplex method which provides an algorithm that involves moving from one vertex to another. The unit has explained the duality theorem which states that for every maximization problem in LP, there is a unique and similar problem of minimization involving the same data which describes the original problem.

2.8

NOTES

KEY TERMS

●

Linear programming: A mathematical technique which involves the allocation of limited resources in an optimal manner on the basis of a given criterion of optimality.

●

Solution: A set of values X1 , X2 ... Xn which satisfies the constraints of the LPP is called its solution.

●

Feasible solution: Any solution to a LPP which satisfies the non-negativity restrictions of the LPP is called its feasible solution.

●

Optimal solution: Any feasible solution which optimizes (Minimizes or Maximizes) the objective function of the LPP is called its optimum solution.

●

Non-degenerate: A non-degenerate basic feasible solution is the basic feasible solution which has exactly m positive Xi (i = 1, 2 ......m), i.e., none of the basic variables are zero.

●

Degenerate: A basic feasible solution is said to degenerate if one or more basic variables are zero.

●

Unbounded solution: If the value of the objective function Z can be increased or decreased indefinitely, such solutions are called unbounded solutions.

●

Basic solution: Given a system of m linear equations with n variables (m < n) any solution which is obtained by solving for m variables keeping the remaining n-m variables zero is called a basic solution.

●

Fundamental duality theorem: It states if either the primal or dual problem has a finite optimal solution, then the other problem also has a finite optimal solution and also, the optimal values of the objective function in both the problems are the same, i.e., Max Z = Min Z.

2.9

Linear Programming

ANSWERS TO ‘CHECK YOUR PROGRESS’

1. Linear programming is a mathematical technique which involves the allocation of limited resources in an optimal manner on the basis of a given criterion of optimality. Self-Instructional Material 73

Operations Research

2. The general formulation of the LPP can be stated as follows: Maximize or Minimize Z – C1X1 + C2X2 + ... + CnXn

NOTES

3. The linear programming problem can be expressed in the matrix form as follows: Maximize or Minimize Z = CX

⎛≤⎞ Subject to ⎜⎜ = ⎟⎟ b ⎜≥⎟ ⎝ ⎠ X≥0 ⎛ x1 ⎞ ⎜ ⎟ ⎜ x2 ⎟ Where x = ⎜ ⎟ , b = ⎜⎜ ⎟⎟ ⎝ xn ⎠ ⎛ a11 ⎜ a21 and A = ⎜ ⎜ ⎜⎜ a ⎝ m1

a12 a22 am 2

⎛ b1 ⎞ ⎜ ⎟ ⎜ b2 ⎟ ⎜ ⎟ , C = (C1C2 – Cn) ⎜⎜ ⎟⎟ ⎝ bm ⎠ a1n ⎞ ⎟ a2 n ⎟ ⎟ ⎟ amn ⎟⎠

4. If the value of the objective function Z can be increased or decreased indefinitely, such solutions are called unbounded solutions. 5. Slack variables are also defined as the non-negative variables which are added in the LHS of the constraint to convert the inequality ‘≤’ into an equation. 6. According to the complementary slackness theorem, (i) If a primal variable is positive, then the corresponding dual constraint is an equation at the optimum and vice versa. (ii) If a primal constraint is a strict inequality then the corresponding dual variable is zero at the optimum and vice versa.

2.10 QUESTIONS AND EXERCISES Short-Answer Questions 1. Briefly describe the graphic and simplex methods of solving a linear programing problem. 2. Write short notes on: (a) Limitations of linear proramming 74 Self-Instructional Material

(b) Dual of a linear programming model

Linear Programming

(c) Goal programming

Long-Answer Questions 1. What is linear programming? What types of problems can linear programming help in solving? What characteristics must a problem have if linear proramming is to be used?

NOTES

2. Why is the simplex method considered superior to the graphic method? Explain. 3.

(a) Maximize: Subject to:

P = 1.4X1 + X2 X1 < 3 2X1 + X2 < 8 3X1 + 4X2 < 24

and

X1 > 0, X2 > 0

4. Solve the following problems by the simplex method: (a) Minimize: Subject to:

3X1 + 2X2 2X1 + 4X2 > 10 4X1 + 2X2 > 10 X2 > 4

and (b)

X1, X2 > 0

Minimize:

Z = 8X1 + 3X2 + 2X3

Subject to:

4X1 + 2X2 + X3 < 8 3X1 + 2X3 < 10 X 1 + X2 + X3 = 4

and

X1, X2, X3 > 0

2.11 FURTHER READING Bierman, Harold, Charles Bonini and W.H. Hausman. 1973. Quantitative Analysis for Business Decisions. Homewood (Illinois): Richard D. Irwin. Gillett, Billy E. 2007. Introduction to Operations Research. New Delhi: Tata McGraw-Hill. Lindsay, Franklin A. 1958. New Techniques for Management Decision Making. New York: McGraw-Hill. Herbert, G. Hicks. 1967. The Management of Organisations. New York: McGrawHill. Massie, Joseph L. 1986. Essentials of Management. New Jersey: Prentice-Hall. Ackoff, R.L. and M.W. Sasieni. 1968. Fundamentals of Operations Research. New York: John Wiley & Sons Inc. Enrick, Norbert Lloyd. 1965. Management Operations Research. New York: Holt, Rinehart & Winston. Self-Instructional Material 75

Transportation Problems

UNIT 3 TRANSPORTATION PROBLEMS NOTES Structure 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12

Introduction Unit Objectives Transportation Problems Test for Optimality Degeneracy in Transportation Problems Unbalanced Transportation Assignment Problems Travelling Salesman Problem Summary Key Terms Answers to ‘Check Your Progress’ Questions and Exercises Further Reading

3.0 INTRODUCTION Transportation problems are one of the subclasses of LPPs in which the objective is to transport various quantities of a single homogenous commodity that are initially stored at various origins, to different destinations in such a way that the transportation cost is minimum. In this unit, you will also learn about the test for optimality and degeneracy in transportation, as well as the assignment problem, the objective of which is to assign a number of origins to an equal number of destinations at minimum cost or maximum profit.

3.1 UNIT OBJECTIVES After going through this unit, you will be able to: ●

Explain and solve transportation problems

●

Perform the optimality test using the MODI method

●

Resolve degeneracy in transportation problems

●

Describe what is unbalanced transportation

●

Solve assignment problems using the Hungarian Method

3.2 TRANSPORTATION PROBLEMS The transportation problem is one of the subclasses of LPPs in which the objective is to transport various quantities of a single homogeneous commodity that are initially stored at various origins, to different destinations in such a way that the transportation Self-Instructional Material 77

Operations Research

NOTES

cost is minimum. To achieve this objective we must know the amount and location of available supplies and the quantities demanded. In addition, we must know the costs that result from transporting one unit of commodity from various origins to various destinations.

Mathematical Formulation Consider a transportation problem with m origins (rows) and n destinations (columns). Let Cij be the cost of transporting one unit of the product from the ith origin to jth destination. ai the quantity of commodity available at origin i, bj the quantity of commodity needed at destination j. xij is the quantity transported from ith origin to jth destination. This following transportation problem can be stated in the tabular form.

The linear programming model representing the transportation problem is given by

subject to the constraints

The given transportation problem is said to be balanced if

ie., if the total supply is equal to the total demand. 78 Self-Instructional Material

Definitions


Feasible Solution Any set of non-negative allocations (xij>0) which satisfies the row and column sum (rim requirement) is called a feasible solution. Basic Feasible Solution A feasible solution is called a basic feasible solution if the number of non-negative allocations is equal to m]n[1, where m is the number of rows and n the number of columns in a transportation table.

NOTES

Non-degenerate Basic Feasible Solution Any feasible solution to a transportation problem containing m origins and n destinations is said to be nondegenerate, if it contains m]n[1 occupied cells and each allocation is in independent positions. The allocations are said to be in independent positions, if it is impossible to form a closed path. Closed path means by allowing horizontal and vertical lines and all the corner cells are occupied.

The allocations in the following tables are not in independent positions.

The allocations in the following tables are in independent positions. Degenerate Basic Feasible Solution If a basic feasible solution contains less than m]n[1 non-negative allocations, it is said to be degenerate.

Optimal Solution Optimal solution is a feasible solution (not necessarily basic) which minimizes the total cost. The solution of a transportation problem can be obtained in two stages, namely initial solution and optimum solution.


Operations Research

Initial solution can be obtained by using any one of the three methods, viz, (i) North west corner rule (NWCR) (ii) Least cost method or matrix minima method

NOTES

(iii) Vogel’s approximation method (VAM) VAM is preferred over the other two methods, since the initial basic feasible solution obtained by this method is either optimal or very close to the optimal solution. The cells in the transportation table can be classified as occupied cells and unoccupied cells. The allocated cells in the transportation table are called occupied cells and the empty cells in a transportation table are called unoccupied cells. The improved solution of the initial basic feasible solution is called optimal solution which is the second stage of solution, that can be obtained by MODI (modified distribution method).

North West Corner Rule Step 1 Starting with the cell at the upper left corner (North west) of the transportation matrix we allocate as much as possible so that either the capacity of the first row is exhausted or the destination requirement of the first column is satisfied i.e., X11 = min (a1,b1). Step 2 If b1>a1, we move down vertically to the second row and make the second allocation of magnitude x22 = min(a2 b1 – x11) in the cell (2, 1). If b1
Solution Since Σai = 34 = Σbj there exists a feasible solution to the transportation problem. We obtain initial feasible solution as follows. The first allocation is made in the cell (1, 1) the magnitude being x11 = min (5, 7) = 5. The second allocation is made in the cell (2, 1) and the magnitude of the allocation is given by x21 = min (8, 7 – 5) = 2. 80 Self-Instructional Material


NOTES

The third allocation is made in the cell (2, 2) the magnitude x22 = min (8 – 2, 9) = 6. The magnitude of fourth allocation is made in the cell (3, 2) given by min (7, 9 – 6) = 3. The fifth allocation is made in the cell (3, 3) with magnitude x33 = min (7 – 3, 14) = 4. The final allocation is made in the cell (4, 3) with magnitude x43 = min (14, 18 – 4) = 14. Hence, we get the initial basic feasible solution to the given T.P. and is given by X11 = 5; X21 = 2; X22 = 6; X32 = 3; X33 = 4; X43 = 14 Total cost = 2 e 5 – 3 e 2 – 3 e 6 – 3 e 4 – 4 e 7 – 2 e 14 =10 – 6 – 18 – 12 – 28 –28 = Rs 102. Example 3.2 Determine an initial basic feasible solution to the following transportation problem using N.W.C.R

Solution The problem is a balanced TP as the total supply is equal to the total demand. Using the steps involved we find the initial basic feasible1 solution as given in the following table Check Your Progress 1. Define transportation problem. 2. What is a non-degenerate basic feasible solution? 3. List the different methods used to find the initial solution of a transportation problem.

The solution is given by X11 = 6; X12 = 8; X23 = 14; X33 = 1; X34 = 4

4. Why is Vogel's Approximation Method preferred over the other methods?

Total Cost 6 × 6 + 4 × 8 + 2 × 9 + 2 × 14 +6 × 1 + 2 × 4 = Rs 128 Self-Instructional Material 81

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NOTES

Least Cost or Matrix Minima Method Step 1 Determine the smallest cost in the cost matrix of the transportation table. Let it be Cij. Allocate xij = min (ai, bj) in the cell (i, j). Step 2 If xij = ai cross off the ith row of the transportation table and decrease bj by ai. Then go to step 3. If xij = bj cross off the jth column of the transportation table and decrease ai by bj. Go to step 3. If xij = ai = bj cross off either the ith row or the jth column but not both. Step 3 Repeat steps 1 and 2 for the resulting reduced transportation table until all the rim requirements are satisfied. Whenever the minimum cost is not unique, make an arbitrary choice among the minima. Example 3.3 Obtain an initial feasible solution to the following TP using Matrix Minima Method.

Solution Since Σai = Σbj = 24, there exists a feasible solution to the TP using the steps in the least cost method, the first allocation is made in the cell (3, 1) the magnitude being x31 = 4. This satisfies the demand at the destination D1 and we delete this column from the table as it is exhausted.

The second allocation is made in the cell (2, 4) with magnitude x24 = min (6, 8) = 6. Since it satisfies the demand at the destination D4, it is deleted from the table. From the reduced table, the third allocation is made in the cell (3, 3) with magnitude x33 = min (8, 6) = 6. The next allocation is made in the cell (2, 3) with magnitude x23 of min (2, 2) = 2. Finally, the allocation is made in the cell (1, 2) with magnitude x12 = min (6, 6)\6. Now all the rim requirements have been satisfied and hence, the initial feasible solution is obtained. The solution is given by x12 = 6, x23 = 2, X24 = 6, X31 = 4, X33 = 6 Since the total number of occupied cells = 5 < m + n – 1. We get a degenerate solution. Total cost = 6 × 2 + 2 × 2 + 6 × 0 + 4 × 0 + 6 × 2 = 12 + 4 + 12 = Rs 28. 82 Self-Instructional Material

Example 3.4 Determine an initial basic feasible solution for the following TP, using the least cost method.


NOTES

Solution Since Σai = Σ bj, there exists a basic feasible solution. Using the steps in least cost method we make the first allocation to the cell (1, 3) with magnitude X13 = min (14, 15) = 14. (as it is the cell having the least cost). This allocation exhaust the first row supply. Hence, the first row is deleted. From the reduced table, the next allocation is made in the next least cost cell (2, 3) which is chosen arbitrarily with magnitude X23 = Min (1, 16) = 1. This exhausts the 3rd column destination. From the reduced table the next least cost cell is (3, 4) for which allocation is made with magnitude min (4, 5) = 4. This exhausts the destination D4 requirement. Delete this fourth column from the table. The next allocation is made in the cell (3, 2) with magnitude X32 = Min (1, 10) = 1. This exhausts the 3rd origin capacity. Hence, the 3rd row is exhausted. From the reduced table the next allocation is given to the cell (2,1) with magnitude X21 = min (6, 15) = 6. This exhausts the first column requirement. Hence, it is deleted from the table. Finally, the allocation is made to the cell (2, 2) with magnitude X22 = Min (9, 9) = 9 which satisfies the rim requirement. These allocation are shown in the transportation table as follows.


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The following table gives the initial basic feasible solution.

NOTES

The solution is given by X13=14; X21=6; X22= 9; X23= 1; X32= 1; X34= 4 Transportation cost = 14 ×1 × 6 × 8 + 9 × 9 + 1 × 2 + 3 × 1 + 4 × 2 = Rs 156.

Vogel’s Approximation Method (VAM) The steps involved in this method for finding the initial solution are as follows. Step 1 Find the penalty cost, namely the difference between the smallest and next smallest costs in each row and column. Step 2 Among the penalties as found in Step(1) choose the maximum penalty. If this maximum penalty is more than one (i.e., if there is a tie) choose any one arbitrarily. Step 3 In the selected row or column as by Step(2) find out the cell having the least cost. Allocate to this cell as much as possible depending on the capacity and requirements. Step 4 Delete the row or column which is fully exhausted. Again, compute the column and row penalties for the reduced transportation table and then go to Step (2). Repeat the procedure until all the rim requirements are satisfied. Note If the column is exhausted, then there is a change in row penalty and vice versa. Example 3.5 Find the initial basic feasible solution for the following transportation problem by VAM.

Solution Since Σai = Σbj = 950 the problem is balanced and there exists a feasible solution to the problem. First, we find the row Σ column penalty PI as the difference between the least and the next least cost. The maximum penalty is 5. Choose the first column arbitrarily. 84 Self-Instructional Material

In this column, choose the cell having the least cost name (1, 1). Allocate to this cell with minimum magnitude (i.e.(250, 200) = 200.) This exhausts the first column. Delete this column. Since a column is deleted, then there is a change in row penalty PII and column penalty PII remains the same. Continuing in this manner, we get the remaining allocations as given in the following table below.


NOTES

Finally, we arrive at the initial basic feasible solution which is shown in the following table.


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There are 6 positive independent allocations which equals to m + n –1 = 3 + 4 – 1. This ensures that the solution is a non-degenerate basic feasible solution. ∴ The transportation cost

NOTES

= 11 × 200 + 13 × 50 + 18 × 175 + 10 × 125 + 13 × 275 + 10 × 125 = Rs 12,075. Example 3.6 Find the initial solution to the following TP using VAM.

Solution Since Σai = Σ bj the problem is a balance TP. Hence, there exists a feasible solution.

Finally, we have the initial basic feasible solution as given in the following table.

There are 6 independent non-negative allocations equal to m]n[1 \3]4[1\6. This ensures that the solution is non-degenerate basic feasible. ∴ The transportation cost = 3 × 45 + 4 × 30 + 1× 25 + 2 × 80 + 4 × 45 + 1 + 75 = 135 + 120 + 25 + 160 + 180 + 75 = Rs 695 86 Self-Instructional Material

3.3 TEST FOR OPTIMALITY Once the initial basic feasible solution has been computed, the next step in the problem is to determine whether the solution obtained is optimum or not. Optimality test can be conducted to any initial basic feasible solution of a TP provided such allocations has exactly m + n – 1 non-negative allocations, where m is the number of origins and n is the number of destinations. Also, these allocations must be in independent positions.


NOTES

To perform this optimality test, we shall discuss the modified distribution method (MODI). The various steps involved in the MODI method for performing optimality test are as follows.

MODI Method Step 1 Find the initial basic feasible solution of a TP by using any one of the three methods. Step 2 Find out a set of numbers ui and vj for each row and column satisfying ui ] vj\cij for each occupied cell. To start with, we assign a number ‘0’ to any row of column having maximum number of allocations. If this maximum number of allocations is more than one, choose any one arbitrarily. Step 3 For each empty (unoccupied) cell, we find the sum ui and vj written in the bottom left corner of that cell. Step 4 Find out for each empty cell the net evaluation value Δij =cij–(ui+vj) and which is written at the bottom right corner of that cell. This step gives the optimality conclusion, (i) If all Δij > 0 (i.e., all the net evaluation value) the solution is optimum and a unique solution exists. (ii) If Δij P 0 then the solution is optimum, but an alternate solution exists. (iii) If at least one Δij < 0 , the solution is not optimum. In this case, we go to the next step, to improve the total transportation cost. Step 5 Select the empty cell having the most negative value of Δij . From this cell we draw a closed path by drawing horizontal and vertical lines with the corner cells occupied. Assign sign ] and [ alternately and find the minimum allocation from the cell having negative sign. This allocation should be added to the allocation having positive sign and subtracted from the allocation having negative sign. Step 6 The previous step yields a better solution by making one (or more) occupied cell as empty and one empty cell as occupied. For this new set of basic feasible allocations, repeat from the step(2) till an optimum basic feasible solution is obtained. Example 3.7 Solve the following transportation problem.


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NOTES

Solution We first find the initial basic feasible solution by using VAM. Since Σai= Σbj the given TP is a balanced one. Therefore, there exists a feasible solution. Finally, we have the initial basic feasible solution as given in the following table.

From this table, we see that the number of non-negative independent allocations is 6 = m+ n + 1 = 3+4+1. Hence, the solution is non-degenerate basic feasible. ∴ The initial transportation cost = 11×13+3×14+4×23+6×17+17×10+18×9 = Rs 711 To find the optimal solution We apply the MODI method in order to determine the optimum solution. We determine a set of numbers ui and vj for each row and column, with ui+vj=cij for each occupied cell. To start with we give u2\0 as the 2nd row has the maximum number of allocation. Now, we find the sum ui and vj for each empty cell and enter at the bottom left corner of that cell. Next, we find the net evaluations Δji = Cij-(ui+vj) for each unoccupied cell and enter at the bottom right corner of that cell. Initial Table


Since all Δij > 0, the solution is optimal and unique. The optimum solution is given by


x14 = 11, x21 = 6, x23 = 3, x24 = 4, x32 = 10, x33 = 9 The minimum transportation cost

NOTES

= 11e13 + 17e6 + 3e14 + 4e23 + 10e17 + 9e18 = Rs 711. Example 3.8 Solve the following transportation problem starting with the initial solution obtained by VAM.

Solution Since Σai = Σbj the problem is a balanced TP. Therefore, there exists a feasible solution.

Finally, the initial basic feasible solution is given as follows:

Since the number of occupied cells = 6 = m – n + 1 and are also independent, there exists a non-degenerate basic feasible solution. The initial transportation cost = 3 e 2 + 3 e 5 + 4 e 4 + 7 e 1 + 6 e 3 + 6 e 1 = Rs 68 To find the optimal solution Applying the MODI method, we determine a set of numbers ui and vj for each row and column, such that ui +vj = cij for each occupied Self-Instructional Material 89

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cell. Since the 3rd now has maximum number of allocations we give number u3 = 0. The remaining numbers can be obtained as follows. c31 = u3 + v1 = 7 = 0 + v1 =7 ⇒ v1 = 7

NOTES

c32 = u3 + v2 = 6 = 0 + v2 = 6 ⇒ v2 = 6 c33 = u3 + v3 = 6 = 0 + v3 = 6 ⇒ v3 = 6 c23 = u2 + v3 = 5 = u2 + 6 = 5 ⇒ u2 = –1 c24 = u2 + v4 = 5 = –1 + v4 + 4 ⇒ v4 = 5 c11 = u1 + v1 = 2 = u1 + 7 = 2 ⇒ u1 = 5 We find the sum ui and vj for each empty cell and enter at the bottom left corner of the cell. Next, we find the net evaluation Δij given by Δij = Cij – (ui + vj) for each empty cell and enter at the bottom right corner of the cell. Initial table

Since all Δij > 0 the solution is optimum and unique. The solution is given by x11 = 3; x23 = 3; x24 = 4 x31 = 1; x32 = 3; x33 = 1 The total transportation cost = 2 × 3 + 3 × 5 + 4 × 4 + 7 × 1 + 6 × 3 + 6 × 1 = Rs 68

3.4 DEGENERACY IN TRANSPORTATION PROBLEMS In a TP, if the number of non-negative independent allocations is less than m]n[1, where m is the number of origins (rows) and n is the number of destinations (columns) there exists a degeneracy. This may occur either at the initial stage or at the subsequent iteration. To resolve this degeneracy, we adopt the following steps. (1) Among the empty cell, we choose an empty cell having the least cost which is of an independent position. If this cell is more than one, choose any one arbitrarily. (2) To the cell as chosen in step (1) we allocate a small positive quantity ∈> 0. The cells containing ∈ are treated like other occupied cells and degeneracy is removed by adding one (more) accordingly. For this modified solution, we adopt the steps involved in MODI method till an optimum solution is obtained. 90 Self-Instructional Material

Example 3.9 Solve the transportation problem for minimization.


NOTES

Solution Since Σ ai\Σ bj the problem is a balanced TP. Hence, there exists a feasible solution. We find the initial solution by North West Corner rule as follows.

Since the number of occupied cells = 5 = m + n – 1 and all the allocations are independent, we get an initial basic feasible solution. The initial transportation cost = 10 e 2 + 4 e 10 + 5 e 1 + 10 e 3 + 1 e 30 = Rs 125 To find the optimal solution (MODI method) We use the previous table to apply the MODI method. We find out a set of numbers ui and vj for which ui + vj = cij, only for occupied cell. To start with, as the maximum number of allocations is 2 in more than one row and column, we choose arbitrarily column 1, and assign a number 0 to this column, i.e., v1 = 0. The remaining numbers can be obtained as follows.


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Initial Table

NOTES

We find the sum of ui and vj for each empty cell and write at the bottom left corner of that cell. Find the net evaluations Δij = cij–(ui+vj) for each empty cell and enter at the bottom right corner of the cell. The solution is not optimum as the cell (3, 1) has a negative Δij value. We improve the allocation by making this cell namely (3, 1) as an allocated cell. We draw a closed path from this cell and assign sign + and − alternately. From the cell having negative sign we find the min. allocation given by min (10, 10) = 10. Hence, we get two occupied cells. (2, 1) (3, 2) becomes empty and the cell (3, 1) is occupied and resulting in a degenerate solution. (Degeneracy in subsequent iteration). Number of allocated cells = 4 < m + n − 1 = 5. We get degeneracy. To resolve we add the empty cell (1, 2) and allocate ε>0. This cell namely (1, 2) is added as it satisfies the two steps for resolving the degeneracy. We assign a number 0 to the first row, namely u1 = 0 we get the remaining numbers as follows.

Next, we find the sum of uj and vj for the empty cell and enter at the bottom left corner of that cell and also the net evaluation Δij = cij– (ui+vj) for each empty cell and enter at the bottom right corner of the cell. I Iteration table


The modified solution is given in the following table. This solution is also optimal and unique as it satisfies the optimality condition that all Δij > 0.


NOTES

x11 = 10: x22 = 15; x33 = 30; x12 = ∈; x31 = 10 T

o t a l

c o s t

=

1

0

×

2

+

2

x ∈ + 15 × 1 + 10 × 1 + 30 × 1

= 75 + 2 ∈ = Rs 75 Example 3.10 Solve the following transportation problem whose cost matrix is given below.

Solution Since Σai = Σbj the problem is a balanced transportation problem. Hence, there exists a feasible solution. We find the initial solution by North west corner rule.

We get the total number of allocated cells = 7 = 4 + 4 – 1. As all the allocations are independent, the solution is a non-degenerate solution. Self-Instructional Material 93

Total transportation cost

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= 1×21+ 5 × 13 + 3 × 12 + 3 × 1 + 12 × 2 + 2 × 2 + 17 × 4 = Rs 221

NOTES

To find the optimal solution (MODI Method) We determine a set of numbers ui and vj for each row and each column with uj + vj = cij for each occupied cell. To start with we give 0 to the third column as it has the maximum number of allocation. Initial Table

We find the sum ui and vj for each empty cell and enter at the bottom left corner Δij = cij -(ui + vj) for each empty cell and enter at the bottom right corner of that cell. The solution o f

t h a t

c e l l .

W

e

f i n d

t h e

n e t

e v

a l u

a t i o n

is not optimum as some of Δij<0. We choose the most negative Δij i.e., 2. There is a tie between the cell (1, 4) and (3, 2) but we choose the cell (3, 2) as it has the least cost. From this cell, we draw a closed path and assign + and − signs alternately and find the minimum allocation from the cell having − sign.


Thus, we get, Min (12, 12) = 12. Hence, one empty cell (3, 2) becomes occupied and two occupied cells (2, 2) (3, 3) becomes empty resulting in degeneracy (Degeneracy in subsequent iteration). By adding and subtracting this minimum allocation we get the modified allocation as given in the following table. For this modified allocations, we repeat the steps in the MODI method.

I Iteraton table


NOTES

The number of allocations = 6 < m + n − 1 = 7. We add the cell (3,3) as it is the least cost empty cell which is of independent position. Give a small quantity ∈> 0. This removes degeneracy. The modified allocation is given in the following table.

The solution is not optimum. The next negative value of Δij = [4. (the cell (1, 4)). The minimum allocation is min. (13, ∈, 17) = ∈. Proceeding in the same manner we have the 2nd iteration table given as follows. II Iteration table

As the solution is not optimum, we improve the solution by using the steps involved in the MODI method. The most negative value of Δij = – 2. Min allocation is min. (13 – ∈, 15, 17 – ∈) = 13 – ∈. Self-Instructional Material 95

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III Iteration table

NOTES

Improve the solution by adding and subtracting the new allocation given by min. (2 + ∈1, 4) = (2]∈). IV Iteration table

Since all ΔijP0 the solution is optimum (alternate solution exists). The solution is given by

Example 3.11 A company has three plants A, B and C, 3 warehouses X, Y, Z. The number of units available at the plants is 60, 70, 80 and the demand at X, Y, Z are 50, 80, 80 respectively. The unit cost of the transportation is given in the following table. 96 Self-Instructional Material


NOTES Find the allocation so that the total transportation cost is minimum. Solution

Since Σ ai = Σbj = 210 the problem is a balanced one. Hence, there exists a feasible solution. Let us find the initial solution by least cost method.

Here, the least cost cell is not unique, i.e., the cells (2, 1) (1, 3) and (3, 2) have the least value 3. So choose the cell arbitrarily. Let us choose the cell (2, 1) and allocate with magnitude min. (70, 50) = 50. This exhausts the first column. Now, delete this column. The reduced transportation table is given by

Continuing in this manner, we finally arrive at the initial solution which is shown in the following table:


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The number of allocated cells is 4 < m + n – 1 = 5, resulting in degeneracy (Initial stage). To remove this degeneracy we add an empty cell (3, 3) whose cost is minimum and is of independent position. Allocate to this cell a small quantity ∈>0. Hence, we have the initial solution given in the following table

NOTES

The total number of occupied cells = 5 = m + n + 1 =5. These are also of independent position. This solution is non-degenerate. The solution is given by X13 = 60, x21 = 50 X23 = 20, X32 = 80 X33 = ∈ The total transportation cost = 3 × 60 + 3 × 50 + 9 × 20 + 3 × 80 + 5 × ∈ = Rs 750 + 5 ∈ = Rs 750 To find the optimal solution We apply the steps involved in the MODI method to the previous table. We find a set of numbers ui and vj for which ui + vj = cij is satisfied for each of the occupied cell. To start with we assign a number 0 to the third column (i.e., v3 = 0) as it has the maximum number of allocations. The remaining numbers are obtained as follows.

Since all Δij > 0 we have obtained an optimum solution. The solution is given by X13 = 60; X21 = 50; X23 = 20; X32 = 80; X33 = ∈ Total transportation cost = 30 × 60 × 3 × 50 × 9 × 20 × 3 × 80 × 5 × ∈ = Rs 750 + 5 ∈ = Rs 750 98 Self-Instructional Material


3.5 UNBALANCED TRANSPORTATION The given TP is said to be unbalanced if Σai ≠ Σbj , i.e., if the total supply is not equal to the total demand.

NOTES

There are two possible cases. Case i :

m

n

i=1

j =1

∑ ai < ∑ b j

If the total supply is less than the total demand, a dummy source (row) is included in the cost matrix with zero cost; the excess demand is entered as a rim requirement for this dummy source (origin). Hence, the unbalanced transportation problem can be converted into a balanced TP. Case ii :

m

n

i=1

j =1

∑ ai > ∑ b j

i.e., the total supply is greater than the total demand. In this case, the unbalanced TP can be converted into a balanced TP by adding a dummy destination (column) with zero cost. The excess supply is entered as a rim requirement for the dummy destination. Example 3.12 Solve the transportation problem when the unit transportation costs, demands and supplies are as given:

Solution Since the total demand Σbj = 215 is greater than the total supply Σai =195 the problem is an unbalanced TP. with cost zero and giving supply equal to 215−195=20 units. Hence, we have the converted problem as follows W

e

c o n v

e r t

t h i s

i n t o

a

b

a l a n c e d

T

P

b

y

i n t r o d

u

c i n g

a

d

u

m

m

y

o r i g

i n

0

4

As this problem is balanced, there exists a feasible solution to this problem. Using VAM we get the following initial solution.


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NOTES

The initial solution to the problem is given by

There are 7 independent non-negative allocations equal to m + n – 1. Hence, the solution is a non-degenerate one. The total transportation cost = 6 × 65 + 5 × 1 + 5 × 30 + 2 × 25 + 4 × 25 + 7 × 45 + 20 × 0 = Rs 1010 To find the optimal solution We apply the steps in the MODI method to the previous table.



NOTES

Since all Δij P 0 the solution is not optimum, we introduce the cell (3,1) as this cell has the most negative value of Δij. We modify the solution by adding and subtracting the min allocation given by min (65, 30, 25). While doing this, the occupied cell (3, 3) becomes empty. I Iteration Table

As the number of independent allocations are equal to m+n–1, we check the optimality. Since all Δij ≥ 0, the solution is optimal and an alternate solution exists as Δ14 = 0. Therefore, the optimum allocation is given by X11 = 40, X12 = 30, X22 = 5, X23 = 50, X31 = 25, X34 = 45, X41 = 20. The optimum transportation cost is = 6 × 40 + 1 × 30 + 5 × 5 + 2 × 50 + 10 × 25 + 7 × 45 + 0 × 20 = Rs 960 Example 3.13 A product is produced by 4 factories F1 F2 F3 and F4. Their unit production cost are Rs 2, 3, 1 and 5 respectively. Production capacity of the factories are 50, 70, 40 and 50 units respectively. The product is supplied to 4 stores S1, S2, S3 and S4, the requirements of which are 25, 35, 105 and 20 respectively. Find the transportation plan such that the total production and transportation cost is minimum. Unit costs of transportation are given as follow. Self-Instructional Material 101

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NOTES

Solution We form the transportation table which consists of both production and transportation costs.

Total capacity = 200 units Total demand = 185 units Therefore, ∑ai > ∑bj. Hence, the problem is unbalanced. We convert it into a balanced one by adding a dummy store S5 with cost 0 and the excess supply is given as the rim requirement to this store, namely (200–185) units.

The initial basic feasible solution is obtained by least cost method. We get the solution containing 8 non-negative independent allocations equals to m + n – 1. So the solution is a non-degenerate solution. The total transportation cost = 4 × 25 + 6 × 5 + 8 × 20 + 10 × 50 + 8 × 20 + 4 × 30 + 13 × 55 + 0 × 15 = Rs 1525 To find the optimal solution We apply MODI method to the previous table as it has m + n – 1 independent non-negative allocations.


Initiable Table


NOTES

The solution is not optimum as the cell (4, 5) is having a negative net evaluation value, i.e., Δ45 = –3. We draw a closed path from this cell and have a modified allocation by adding and subtracting the allocation min (35, 20) = 20. This modified allocation is given in the following table. I Iteration Table

Since all the values of Δij P 0 the solution is optimum, but an alternate solution exists. The optimum solution or the transportation plan is given by X11 = 25 units

X32 = 30 units

X12 = 5 units

X43 = 15 units

X13 = 20 units

X44 = 20 units

X23 = 70 units

X45 = 15 units

This is the surplus capacity that are not transported which are manufactured in factory F4. The optimum production with transportation cost = 4 × 25 + 6 × 5 + 8 × 20 + 10 × 70 + 4 × 30 + 8 e 20 + 13 × 15 + 0 × 15 = Rs 1465 Maximization case in transportation problem In this, the objective is to maximize the total profit for which the profit matrix is given. For this, first we have to convert the maximization problem into minimization by subtracting all the Self-Instructional Material 103

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NOTES

elements from the highest element in the given transportation table. This modified minimization problem can be solved in the usual manner. Example 3.14 There are three factories A, B and C which supply goods to four dealers D1, D2, D3 and D4. The production capacities of these factories are 1000, 700 and 900 units per month respectively. The requirements from the dealers are 900, 800, 500 and 400 units per month respectively. The per unit return (excluding transportation cost) are Rs 8, Rs 7 and Rs 9 at the three factories. The following table gives the unit transportation costs from the factories to the dealers.

Determine the optimum solution to maximize the total returns. Solution Profit\return–transportation cost. With this we form a transportation table with profit.

Profit matrix

This profit matrix is converted into its equivalent loss matrix by subtracting all the elements from the highest element, namely 8. Hence, we have the following loss matrix.

We use VAM to get the initial basic feasible solution.



NOTES

The initial solution is given in the following table:

Since the number of allocated cell = 5 < m + n – 1 = 6, the solution is degenerate. To resolve this degeneracy we add an empty cell (1, 3) by allocating a non-negative quantity ∈. This cell is the least cost cell and is of independent position. The initial basic feasible solution is given as follows: Initial Table

Number of allocations = 6 = m + n – 1 and the 6 allocations are in independent position. Hence, we can perform the optimality test using MODI method. Self-Instructional Material 105

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NOTES

A

200

B

700

800

4

1

2

D3 2 ∈

4

2 2

1

1

2

2

4 1

3 3

3

0 0

2 4

ui

D4 2

6

3

C ui

D2 2

2 1 500

400

2

1

0 2 –1

Since all the net evaluation Δji > 0 are non-negative the initial solution is optimum. The optimum distribution is A → D1 = 200 units A → D2 = 200 units A → D3 = ∈ units B → D1 = 700 units C → D3 = 500 units A → D4 = 400 units Total profit or the Max. return = 200 × 6 + 6 × 800 + 4 × 700 + 7 × 500 + 8 × 400 = Rs 15,500. Example 3.15 Solve the following transportation problem to maximize the profit.

Solution Since the given problem is to maximize the profit, we convert this into loss matrix and minimize it. For converting it into minimization type, we subtract all the elements from the highest element 90. Hence, we have the following loss matrix.

Since Σai = Σbj there exists a feasible solution and is obtained by VAM.



NOTES

The initial basic feasible solution is given in the following table.

As the number of independent allocated cells = 6 = m + n – 1 the solution is non-degenerate solution.

Optimality test using MODI method

Since all the net evaluation Δij > 0 the solution is optimum and unique. The optimum solution is given by X12 = 23; X21 = 6; X24 = 30; X31 = 17; X33 = 16 The optimum profit = 23 × 51 + 6 × 80 + 42 × 8 + 81 × 30 + 90 × 17 + 66 × 16 = Rs 7005 Self-Instructional Material 107

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NOTES

3.6 ASSIGNMENT PROBLEMS Suppose there are n jobs to be performed and n persons are available for doing these jobs. Assume that each person can do one job at a time, though with varying degrees of efficiency. Let cij be the cost if the ith person is assigned to the jth job. The problem is to find an assignment (which job should be assigned to which person on a oneone basis) so that the total cost of performing all the jobs is minimum. Problems of this type are known as assignment problems. The assignment problem can be stated in the form of a non cost matrix [cij] of real numbers as given in the following table. Jobs j...

n

1

2

3...

1 c11

c12

c13

c1j

c1n

2 3

c21 c32

c22 c32

c23 c23

c2j c3j

c2n c3n

i

ci 1

ci 2

ci2

cij

cin

c n2

cn3...

cnj...

cnn

...

Persons

n

c n1

Mathematical Formulation of the Assignment Problem Mathematically, the assignment problem can be stated as n

n

i=1

j =1

∑ ∑c x

ij ij

i = 1, 2,...n

j = 1,2,...n

Subject to the restrictions ⎧1 if the i th person is assigned j th job xij = ⎨ ⎩ 0 if not n

∑x

ij

= 1 (one job is done by the ith person)

j =1 n

and

∑x

ij

= 1 (only one person should be assigned the jth job)

i =1

Where xij denotes that the jth job is to be assigned to the ith person.


Difference between Transportation Problem and Assignment Problem Transportation problem 1. No. of sources and number of destinations need not be equal. Hence, the cost matrix is not necessarily a square matrix.

Assignment problem Since assignment is done on one-toone basis, the number of sources and the number of destinations are equal. Hence, the cost matrix must be a square matrix.

2 xij, the quantity to be transported from ith origin to jth destination can take any possible positive values, and satisfies the rim requirements.

xij, the jth job is to be assigned to the ith person and can take either the value l or 0.

3. The capacity and the requirement value is equal to ai and bj for the ith source and jth destinations (i=1, 2m j=1,2… n)

The capacity and the requirement value is exactly 1 i.e., for each source of each destination the capacity and the requirement value is exactly 1.

4. The problem is unbalanced if the total supply and the total demand are not equal.

The problem is unbalanced if the cost matrix is not a square matrix.


NOTES

Hungarian Method Procedure The solution of an assignment problem can be arrived using the Hungarian Method. The steps involved in this method are as follows. Step 1 Prepare a cost matrix. If the cost matrix is not a square matrix then add a dummy row (column) with zero cost element. Step 2 Subtract the minimum element in each row from all the elements of the respective rows. Step 3 Further modify the resulting matrix by subtracting the minimum element of each column from all the elements of the respective columns. Thus, obtain the modified matrix. Step 4 Then, draw minimum number of horizontal and vertical lines to cover all zeroes in the resulting matrix. Let the minimum number of lines be N. Now, there are two possible cases. Case i If N = n, where n is the order of matrix, then an optimal assignment can be made. So make the assignment to get the required solution. Case ii If N < n, then proceed to step 5. Step 5 Determine the smallest uncovered element in the matrix (element not covered by N lines). Subtract this minimum element from all uncovered elements and add the same element at the intersection of horizontal and vertical lines. Thus, the second modified matrix is obtained. Step 6 Repeat step (3) and (4) until we get the case (i) of Step 4. Step 7 (To make zero assignment) Examine the rows successively until a rowwise exactly single zero is found. Circle (O) this zero to make the assignment. Then mark a cross (×) over all zeroes if lying in the column of the circled zero, showing Self-Instructional Material 109

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that they can not be considered for future assignment. Continue in this manner until all the zeroes have been examined. Repeat the same procedure for the column also. Step 8 Repeat the step 6 successively until one of the following situation arises

NOTES

— (i) If no unmarked zero is left, then the process ends. (ii) If there lies more than one of the unmarked zero in any column or row, then circle one of the unmarked zeroes arbitrarily and mark a cross in the cells of remaining zeroes in its row or column. Repeat the process until no unmarked zero is left in the matrix. Step 9 Thus, exactly one marked circled zero in each row and each column of the matrix is obtained. The assignment corresponding to these marked circled zeroes will give the optimal assignment. Example 3.16 Using the following cost matrix, determine (a) the optimal job assignment (b) the cost of assignments.

Solution Select the smallest element in each row and subtract this smallest element from all the elements in its row.

Subtract the minimum element from each column and subtract this element from all the elements in its column. With this, we get the first modified matrix.

In this modified matrix, we draw the minimum number of lines to cover all zereo (horizontal or vertical). 110 Self-Instructional Material


NOTES

The number of lines drawn to cover all zeroes is 4 = N. The order of matrix is n = 5. Hence, N < n. Now we get the second modified matrix by subtracting the smallest uncovered element from the remaining uncovered elements and add it to the element at the point of intersection of lines.

The number of lines drawn to cover all zeroes = N = 5. The order of matrix is n = 5. Hence, N = n. Now, we determine the optimum assignment.

Assignment

The first row contains more than one zero. So proceed to the second row. It has exactly one zero. The corresponding cell is (B, 4). Circle this zero, thus making an assignment. Mark (×) for all other zeroes in its column. Showing that they cannot be used for making other assignments. Now, row 5 has a single zero in the cell (E, 3). Make an assignment in this cell and cross the second zero in the third column. Self-Instructional Material 111

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Now, row 1 has a single zero in the column 2, i.e., in the cell (A, 2). Make an assignment in this cell and cross the other zeroes in the second

NOTES

column. This leads to a single zero in column 1 of the cell (D, 1), second. Make an assignment in this cell and cross the other zeroes in the fourth. row. Finally, we have a single zero left in the third row making an assignment in the cell (C, 5). Thus, we have the following assignment. Optimal assignment and optimum cost of assignment. Job

Mechanic

Cost

1

D

3

2

A

3

3

E

9

4

B

2

5

C

4 Rs 21

Therefore 1→D, 2→A, 3→E, 4→B, 5→C with minimum cost Rs 21. Example 3.17 A company has 5 jobs to be done on five machines. Any job can be done on any machine. The cost of doing the jobs in different machines are given below. Assign the jobs for different machines so as to minimise the total cost.

Solution We form the first modified matrix by subtracting the minimum element from all the elements in the respective row and same with respective column.

Since each column has the minimum element 0. We have the first modified matrix. Now, we draw the minimum number of lines to cover all zeroes.



NOTES

The number of lines drawn to cover all zeroes is N = 4 < the order of matrix is n = 5. We find the second modified matrix by subtracting the smallest uncovered element from all the uncovered elements and add to the element which is in the point of intersection of lines.

The number of lines drawn to cover all zeroes = 5. This is the order of matrix. Hence, we can form an assignment.

Assignment

All the five jobs have been assigned to 5 different machines. Here, the optimal assignment is. Job

Machine

1

B

2

E

3

C

4

A

5

D

Minimum (Total cost) = 8+12+4+6+12 = Rs 42 Self-Instructional Material 113

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Example 3.18 Four different jobs can be done on four different machines and take down time costs are prohibitively high for change overs. The following matrix gives the cost in rupees of producing job i on machine j:

NOTES

How should the jobs be assigned to the various machines so that the total cost is minimized? Solution We form a first modified matrix by subtracting the least element in the respective row and respective column.

Since the third column has no zero element, we subtract the smallest element 4 from all the elements.

Now we draw minimum number of lines to cover all zeroes. The number of lines drawn to cover all zeroes = 3 which is less than the order of matrix\4. Hence, we form the second modified matrix, by subtracting the smallest uncovered element from all the uncovered elements and adding to the element which is in the point of intersection of lines. 114 Self-Instructional Material


NOTES

N=3=n=4

N=4=n=4

Hence, we can make an assignment. Since no rows and no columns have single zero, we have different assignment (Multiple solution). Optimal assignment Job

Machine

J1

M4

J2

M2

J3

M1

J4

M3

Minimum (Total cost) 6+5+4+8 = Rs 23

Alternate solution J1 → M1; J2→ M2 = J3→ M3; J4→M4. Minimum (Total cost) 5+5+10+3 = Rs 23 Example 3.19 Solve the following assignment problem in order to minimize the total cost. The following cost matrix given gives the assignment cost when different operators are assigned to various machines. Self-Instructional Material 115

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NOTES

Solution We form the first modified matrix by subtracting the least element from all the elements in the respective rows and then in the respective columns.

Since each column has the minimum element 0, the first modified matrix is obtained. We draw the minimum number of lines to cover all zeroes. The number of lines drawn to cover all zeroes = 4 = the order of matrix = 5. Hence, we form the second modified matrix by subtracting the smallest uncovered element from the remaining uncovered elements and adding to the element which is in the point of intersection of lines.

N = 5, i.e., the number of lines drawn to cover all zeroes = order of matrix. Hence, we can make assignment. The optimum assignment is Operators

Machines

I

D

II

A

III

C

IV

E

V

B

The optimum cost is given by 116 Self-Instructional Material

25+25+22+24+26 = Rs 122

Unbalanced Assignment problem Any assignment problem is said to be unbalanced if the cost matrix is not a square matrix, i.e., the number of rows and the number of columns are not equal. To make it balanced, we add a dummy row or a dummy column with all the entries as zero.


NOTES

Example 3.20 There are four jobs to be assigned to the machines. Only one job could be assigned to one machine. The amount of time in hours required for the jobs in a machine are given in the following matrix.

Find an optimum assignment of jobs to the machines to minimize the total processing time and also, find for which machine no job is assigned. What is the total processing time to complete all the jobs? Solution Since the cost matrix is not a square matrix, the problem is unbalanced. We add a dummy job 5 with corresponding entries zero.

Modified matrix

We subtract the smallest element from all the elements in the respective row.

Check Your Progress

Since each column has minimum element as zero, we draw the minimum number of lines to cover all zeroes.

5. When does degeneracy occur in transportation problems? 6. When do you say that a TP is unbalanced? 7. Cite one difference between a transportation problem and an assignment problem. 8. What is an unbalanced assignment problem? Self-Instructional Material 117

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The number of lines to cover all zeroes = 4 < the order of matrix. We form the second modified matrix by subtracting the smallest uncovered element from the remaining uncovered elements and adding to the element at the point of intersection of lines.

NOTES

Here the number of lines drawn to cover all zeroes = 5 = Order of matrix. Therefore, we can make the assignment

Optimum assignment Job

Machine

1

A

2

C

3

D

4

E

For machine E no job is assigned. Optimum (minimum) cost = 4 + 11 + 1 + 6 = 22. Example 3.21 A company has 4 machines to do 3 jobs. Each job can be assigned to one and only one machine. The cost of each job on each machine is given as follows. Determine the job assignments which will minimize the total cost.


Solution Since the cost matrix is not a square matrix we add a dummy row D with all the elements 0.


NOTES

Subtract the minimum element in each row from all the elements in its row.

Since each column has minimum element we draw minimum number of lines to cover all zeroes.

∴the number of lines drawn to cover all zeroes = 2 < the order of matrix, we form a second modified matrix.

Here, N = 3 < n = 4. Again, we subtract the smallest uncovered element from all the uncovered elements and add to the element at the point of intersection

Here, N = 4 = n. Hence, we make an assignment. Self-Instructional Material 119

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Assignment

NOTES

Since D is a dummy job, machine Z has assigned no job. Therefore, optimum cost = 18 + 13 + 19 = Rs 50

Maximization in Assignment problem In this, the objective is to maximize the profit. To solve this, we first convert the given profit matrix into the loss matrix by subtracting all the elements from the highest element of the given profit matrix. For this converted loss matrix, we apply the steps in Hungarian method to get the optimum assignment. Example 3.22 The owner of a small machine shop has four mechanics available to assign jobs for the day. Five jobs are offered with expected profit for each mechanic on each job which are as follows:

Find by using the assignment method, the assignment of mechanics to the job that will result in a maximum profit. Which job should be declined? Solution The given profit matrix is not a square matrix as the number of jobs is not equal to the number of mechanics. Hence, we introduce a dummy mechanic 5 with all the elements 0.

Now we convert this profit matrix into loss matrix by subtracting all the elements from the highest element 111. 120 Self-Instructional Material


NOTES

We subtract the smallest element from all the elements in the respective row.

Since each column has the minimum element as zero, we draw the minimum number of lines to cover all zeroes.

Here, the number of lines drawn to cover all zeroes = N = 4 is less than the order of matrix. We form the second modified matrix by subtracting the smallest uncovered element from the remaining uncovered elements and adding to the element which is at the point of intersection of lines.

Here, N = 5 = n (the order of matrix). We make the assignment. Self-Instructional Material 121

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NOTES

The optimum assignment is Job A B C D E

Mechanic 5 2 3 1 4

Since the fifth mechanic is a dummy, job A is assigned to the fifth mechanic, this job is declined. The maximum profit is given by 84 + 111 + 101 + 80 = Rs 376 Example 3.23 A marketing manager has 5 salesmen and there are 5 sales districts. Considering the capabilities of the salesmen and the nature of districts, the estimates made by the marketing manager for the sales per month (in 1000 rupees) for each salesmen in each district would be as follows.

Find the assignment of salesmen to the districts that will result in the maximum sales. Solution We are given the profit matrix. To maximize the profit, first we convert it into loss matrix which can be minimized. To convert it in to loss matrix, we subtract all the elements from the highest element 41. Subtract the smallest element from all the elements in the respective rows and colums to get the first modified matrix.



NOTES

We now draw the minimum number of lines to cover all zeroes.

N=4
Again, N = 4 < n = 5. Repeat the previous step.

N = 5 = n = 5. Hence, we make the assignment. Self-Instructional Material 123

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Assignment

NOTES

Since no row or column has single zero, we get a multiple solution. (i) The optimum assignment is: 1 → B, 2 → A, 3 → E, 4 → C, 5 → D With maximum profit (38]40]37]41]35)\Rs 191 (ii) The optimum assignment is: 1 → B, 2 → A, 3 → E, 4 → D, 5 → C Maximum profit (38]40]37]36]40)\Rs 191

3.7 TRAVELLING SALESMAN PROBLEM Assume a salesman has to visit n cities. He wishes to start from a particular city, visit each city once and then return to his starting point. His objective is to select the sequence in which the cities are visited in such a way that his total travelling time is minimized. To visit 2 cities (A and B, there is no choice. To visit 3 cities we have 2! possible routes. For 4 cities we have 3! possible routes. In general, to visit n cities there are (n – 1) ! possible routes.

Mathematical Formulation Let Cij be the distance or time or cost of going from city i to city j. the decision variable Xij be 1 if the salesman travels from city i to city j and otherwise 0. The objective is to minimize the travelling time.

Subject to the constraints


and subject to the additional constraint that Xij is so chosen that no city is visited twice before all the cities are completely visited. I n

p

a r t i c u

l a r ,

g

o i n g

f r o m


i directly to j is not permitted. Which means Cij = ∞,

when i = j. In travelling salesman problem we cannot choose the element along the diagonal and this can be avoided by filling the diagonal with infinitely large elements.

NOTES

The travelling salesman problem is very similar to the assignment problem except that in the former care, there is an additional restriction, that Xij is so chosen that no city is visited twice before the tour of all the cities is completed. Solution Treat the problem as an assignment problem and solve it using the same procedures. If the optimal solution of the assignment problem satisfies the additional constraint, then it is also an optimal solution of the given travelling salesman problem. If the solution to the assignment problem does not satisfy the additional restriction then after solving the problem by assignment technique we use the method of enumeration. Example 3.24 A travelling salesman has to visit 5 cities. He wishes to start from a particular city, visit each city once and then return to his starting point. Cost of going from one city to another is shown below. You are required to find the least cost route.

Solution First we solve this problem as an assignment problem.

Subtract the minimum element in each row from all the elements in its row.

Subtract the minimum element in each column from all the elements in its column.


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NOTES

We have the first modified matrix. Draw minimum number of lines to cover all zeroes.

N = 4 0 n = 5 Subtract the smallest uncovered element from all the uncovered elements and add to the element which is in the point of intersection of lines. Hence, we get the second modified matrix.

N = 5, = n = 5. We make the assignment.

Assignment

As the salesman should go from A to E and then come back to A without covering B, C, D which is contradicting the fact that no city is visited twice before all the cities are visitied. Hence, we obtain the next best solution by bringing the next minimum nonzero element, namely 4. 126 Self-Instructional Material


NOTES

A → B, B → C, C → D, D → E, E → A Since all the cities have been visited and no city is visited twice before completing the tour of all the cities, we have an optimal solution to the travelling salesman. The least cost route is A → B → C → D → E → A. Total cost = 4 + 6 + 8 + 10 + 2 = Rs 30 Example 3.25 A machine operator processes five types of items on his machine each week and must choose a sequence for them. The set-up cost per change depends on the items presently on the machine and the set-up to be made according to the following table.

If he processes each type of item only once in each week, how should he sequence the items on his machine in order to minimize the total set-up cost? Solution Reduce the cost matrix and make assignments in rows and columns having single row.

Modify the matrix by subtracting the least element from all the elements in its row and also in its column.


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NOTES

Here, N = 4 n = 5, i.e., N < n. Subtract the smallest uncovered element from all the uncovered elements and add to the element which is at the point of intersection of lines and get the reduced second modified matrix.

N= 5 = n = 5. We make the assignment.

Assignment

We get the solution A → B → D → A. This schedule provides the required solution as each item is not processed once in a week. Hence, we make a better solution by considering the next smallest non-zero element by considering 1.

A → E, E → C, C → B, B → D, D → A, i.e., A → E → C → B → D → A The total set-up cost comes to Rs 21. 128 Self-Instructional Material

3.8 SUMMARY In this unit, you have learned that in order to achieve the minimum cost in transportation problems, we must know the costs that result from transporting one unit of a commodity from various origins to various destinations. The optimal solution is a feasible solution which minimizes the total cost. The unit has described the MODI method for determining the optimal solution. You have also learned that degeneracy may occur either at the initial stage or at any subsequent iteration. The unit has explained that a transportation problem is said to be unbalanced if the total supply is not equal to the total demand, whereas in an assignment problem the problem is unbalanced if the cost matrix is not a square matrix. In the travelling salesman problem, the objective of the salesman is to select a sequence in which the cities are visited in such a way that his total travelling time is minimized.


NOTES

3.9 KEY TERMS ●

Transportation problem: A subclasses of LPPs in which the objective is to transport various quantities of a single homogeneous commodity that are initially stored at various origins, to different destinations in such a way that the transportation cost is minimum.

●

Basic feasible solution: A feasible solution in which the number of nonnegative allocations is equal to m]n[1, where m is the number of rows and n the number of columns in a transportation table.

●

Non-degenerate basic feasible solution: Any feasible solution to a transportation problem containing m origins and n destinations is said to be non-degenerate, if it contains m]n[1 occupied cells and each allocation is in independent positions.

●

Degenerate basic feasible solution: If a basic feasible solution contains less than m]n[1 non negative allocations, it is said to be degenerate.

●

Optimal solution: A feasible solution (not necessarily basic) which minimizes the total cost. The solution of a transportation problem can be obtained in two stages, namely initial solution and optimum solution.

3.10 ANSWERS TO ‘CHECK YOUR PROGRESS’ 1. The transportation problem is one of the subclasses of LPPs in which the objective is to transport various quantities of a single homogeneous commodity that are initially stored at various origins to different destinations in such a way that the transportation cost is minimum. 2. Any feasible solution to a transportation problem containing m origins and n destinations is said to be non-degenerate, if it contains m]n[1 occupied cells and each allocation is in independent positions. Self-Instructional Material 129

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3. Initial solution can be obtained by using any one of the three methods, viz, (i) North west corner rule (NWCR) (ii) Least cost method or matrix minima method (iii) Vogel’s approximation method (VAM)

NOTES

4. VAM is preferred over the other two methods since the initial basic feasible solution obtained by this method is either optimal or very close to the optimal solution. 5. In a TP, if the number of non-negative independent allocations is less than m + n – 1, where m is the number of origins (rows) and n is the number of destinations (columns) there exists a degeneracy. This may occur either at the initial stage or at subsequent iteration. 6. The given TP is said to be unbalanced if Sai ≠ Sbj i.e., if the total supply is not equal to the total demand. 7.

Transportation problem No. of sources and number of destinations need not be equal. Hence, the cost matrix is not necessarily a square matrix.

Assignment problem Since assignment is done on one one basis, the number of sources and the number of destinations are equal. Hence, the cost matrix must be a square matrix.

8. Any assignment problem is said to be unbalanced if the cost matrix is not a square matrix, i.e., the number of rows and the number of columns are not equal. To make it balanced, we add a dummy row or a dummy column with all the entries as zero.

3.11 QUESTIONS AND EXERCISES Short-Answer Questions 1. What do you understand by transportation model? 2. Define feasible solution, basic solution, non-degenerate solution, optimal solution in a transportation problem. 3. Give the mathematical formulation of a TP. 4. Give the mathematical formulation of an assignment problem.

Long-Answer Questions 1. Explain the following briefly with examples: I. North West Corner Rule II. Least Cost Method III. Vogel’s Approximation Method 2. Explain degeneracy in a TP. Describe a method to resolve it. 3. What do you mean by an unbalanced TP? Explain how to convert an unbalanced T.P. into a balanced one. 4. Explain an algorithm to solving a transportation problem. 130 Self-Instructional Material

5. Describe an assignment problem giving a suitable example.


6. Explain the difference between a transportation problem and an assignment problem. 7. How can you maximize an objective function in the assignment problem? Explain the nature of I in traveling salesman problem and give its mathematical formulation.

NOTES

3.12 FURTHER READING Bierman, Harold, Charles Bonini and W.H. Hausman. 1973. Quantitative Analysis for Business Decisions. Homewood (Illinois): Richard D. Irwin. Gillett, Billy E. 2007. Introduction to Operations Research. New Delhi: Tata McGraw-Hill. Lindsay, Franklin A. 1958. New Techniques for Management Decision Making. New York: McGraw-Hill. Herbert, G. Hicks. 1967. The Management of Organisations. New York: McGrawHill. Massie, Joseph L. 1986. Essentials of Management. New Jersey: Prentice-Hall. Ackoff, R.L. and M.W. Sasieni. 1968. Fundamentals of Operations Research. New York: John Wiley & Sons Inc. Enrick, Norbert Lloyd. 1965. Management Operations Research. New York: Holt, Rinehart & Winston.


Decision-Making

UNIT 4 DECISION-MAKING Structure 4.0 Introduction 4.1 Unit Objectives 4.2 Decision-Making Environment

NOTES

4.2.1 Decision-Making under Certainty 4.2.2 Decision-Making under Uncertainty 4.2.3 Decision-Making under Risk

4.3 Decision Tree Analysis 4.4 Integer Programming and Dynamic Programming: Concepts and Advantages 4.4.1 4.4.2 4.4.3 4.4.4 4.4.5 4.4.6

4.5 4.6 4.7 4.8 4.9

4.0

Importance of Integer Programming Problems Applications of Integer Programming Decision Tree and Bellman’s Principle of Optimality Characteristics of DPP Dynamic Programming Algorithm Advantages of DPP


INTRODUCTION

Decision-making is an everyday process in life. It is the major job of a manager too. A decision taken by a manager has far-reaching consequences on the business. Right decisions will have a salutary effect and wrong ones may prove to be disastrous. Decisions may be classified into two categories, tactical and strategic. Tactical decisions are those which affect the business in the short run. Strategic decisions are those which have a long-term effect on the course of business. These days, in every organization—large or small, the person at the top has to take the crucial decisions, knowing that certain events beyond his control may occur and make him regret the decision. He is uncertain as to whether or not these unfortunate events will happen. In such situations the best possible decision can be made with the use of statistical tools which try to minimize the degree to which the person is likely to regret the decision he takes for a particular problem. The problem under study may be represented by a model in terms of the following elements: (i) The decision-maker The decision-maker is charged with the responsibility of taking a decision; he has to select one from a set of possible courses of action. (ii) Acts These are the alternative courses of action or strategies that are available to the decision-maker. The decision involves a selection among two or more alternative courses of action. The problem is to choose the best of these alternatives to achieve an objective. Self-Instructional Material 133

(iii) Events These are the occurrences which affect the achievement of the objectives. They are also called states of nature or outcomes. The events constitute a mutually exclusive and exhaustive set of outcomes which describe the possible behaviour of the environment in which the decision is made. The decision-maker has no control over which event will take place and can only attach a subjective probability of occurence of each.

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NOTES

(iv) Pay-off table It represents the economics of a problem, i.e., the revenue and costs associated with any action with a particular outcome. It is an ordered statement of profit or costs resulting under a given situation. A pay-off can be interpreted as the outcome in quantitative form if the decision-maker adopts a particular strategy under a particular state of nature. (v) Opportunity loss table An opportunity loss is the loss incurred because of the failure to take the best possible action. Opportunity losses are calculated separately for each state of nature that might occur. Given the occurence of a specific state of nature, we can determine the best possible act. For a given state of nature, the opportunity loss of an act is the difference between the pay-off of that act and the pay-off for the best act that could have been selected.

4.1

UNIT OBJECTIVES

After going through this unit, you will be able to: ● ●

● ● ● ●

4.2

Analyse the nature of a decision-making environment Understand the various choices available for decision-making under uncertainty Describe how decisions are taken under risk Describe the procedure of decision tree analysis Understand the importance and applications of integer programming Describe the characteristic features of dynamic programming problems

DECISION-MAKING ENVIRONMENT

In any decision problem, the decision-maker is concerned with choosing from among the available alternative courses of action, the one that yields the best result. If the consequences of each choice are known with certainty, the decision-maker can easily make decisions. But in most of real life problems, the decision-maker has to deal with situations where uncertainty of the outcomes prevail. The decision-making problems can be discussed under the following heads on the basis of their environments: 1. Decision-making under certainty 2. Decision-making under uncertainty 3. Decision-making under risk 4. Decision-making under conflict 134 Self-Instructional Material

Decision-Making

4.2.1 Decision-Making under Certainty In this case, the decision-maker knows with certainty the consequences of every alternative or decision choice. The decision-maker presumes that only one state of nature is relevant for his purpose. He identifies this state of nature, takes it for granted and presumes complete knowledge as to its occurence.

NOTES

4.2.2 Decision-Making under Uncertainty When the decision-maker faces multiple states of nature but he has no means to arrive at probability values to the likelihood of occurence of these states of nature, the problem is a decision problem under uncertainty. Such situations arise when a new product is introduced in the market or a new plant is set up. In business, there are many problems of this ‘nature’. Here, the choice of decision largely depends on the personality of the decision-maker. The following choices are available before the decision-maker in situations of uncertainty. (a) Maximax Criterion (b) Minimax Criterion (c) Maximin Criterion (d) Laplace Criterion (Criterion of equally likelihood) (e) Hurwicz Alpha Criterion (Criterion of Realism) ( a )

The maximax decision criterion (criterion of optimism) The term ‘maximax’ is an abbreviation of the phrase maximum of the maximums, and an adventurous and aggressive decision-maker may choose to take the action that would result in the maximum pay-off possible. Suppose for each action there are three possible pay-offs corresponding to three states of nature as given in the following decision matrix: States of nature S1 S2 S3

A1 220 160 140

Decisions A2 180 190 170

A3 100 180 200

Maximum under each decision are (220, 190, 200). The maximum of these three maximums is 220. Consequently, according to the maximax criteria the decision to be adopted is A1. (b) The minimax decision criterion Minimax is just the opposite of maximax. Application of the minimax criteria requires a table of losses instead of gains. The losses are the costs to be incurred or the damages to be suffered for each of the alternative actions and states of nature. The minimax rule minimizes the maximum possible loss for each course of action. The term ‘minimax’ is an abbreviation of the phrase minimum of the maximum.Under each of the various actions, there is a maximum loss and the action that is associated with the minimum of the various maximum losses is the action to be taken according to the minimax criterion. Suppose the loss table is Self-Instructional Material 135

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Decisions

State of nature

A1

A2

A3

S1 S2

0 3

4 0

10 6

S3

18

14

0

NOTES

It shows that the maximum losses incurred by the various decisions A1

A2

18 14

A3 10

and the minimum among three maximums is 10 which is under action A3. Thus, according to minimax criterion, the decision-maker should take action A3. (c) The maximin decision criterion (criterion of pessimism) The maximin criterion of decision-making stands for choice between alternative courses of action assuming pessimistic view of nature. Taking each act in turn, we note the worst possible results in terms of pay-off and select the act which maximizes the minimum pay-off. Suppose the pay-off table is State of nature S1 S2 S3 S4

A1 –80 –30 30 75

Action A2 –60 –10 15 80

A3 –20 –2 7 25

Minimum under each decision are respectively –80 –60 –20 The action A3 is to be taken according to this criterion because it is the maximum among minimums. (d) Laplace criterion As the decision-maker has no information about the probability of occurrence of various events, he makes a simple assumption that each probability is equally likely. The expected pay-off is worked out on the basis of these probabilities. The act having maximum expected pay-off is selected. Example 4.1: Events


Act A1

A2

A3

E1

20

12

25

E2

25

15

30

E3

30

20

22

Solution: We associate equal probability for each event say, 1/3. Expected pay-offs are: 1⎞ ⎛ 1⎞ ⎛ 1 ⎞ 75 ⎛ A1 → ⎜ 20 × ⎟ + ⎜ 25 × ⎟ + ⎜ 30 × ⎟ = = 25 3⎠ ⎝ 3⎠ ⎝ 3⎠ 3 ⎝ 1⎞ ⎛ 1⎞ ⎛ 1 ⎞ 47 ⎛ A 2 → ⎜12 × ⎟ + ⎜15 × ⎟ + ⎜ 20 × ⎟ = = 15.67 3⎠ ⎝ 3⎠ ⎝ 3⎠ 3 ⎝

Decision-Making

NOTES

1⎞ ⎛ 1⎞ ⎛ 1 ⎞ 77 ⎛ A3 → ⎜ 25 × ⎟ + ⎜ 30 × ⎟ + ⎜ 22 × ⎟ = = 25.67 3⎠ ⎝ 3⎠ ⎝ 3⎠ 3 ⎝

Since A3 has the maximum expected pay-off, A3 is the optimal act. (e) Harwicz alpha criterion This method is a combination of maximum criterion and maximax criterion. In this method, the decision-maker’s degree of optimism is represented by α ,the coefficient of optimism. α varies between 0 and 1. When α = 0, there is total pessimism and when α = 1, there is total optimism. we find D1, D2, D3, etc. connected with all strategies where Di = a Mi + (1 – a) mi where Mi is the maximum pay-off of ‘i’ the strategy and mi is the minimum pay-off of ‘i’ th strategy. The strategy with highest of D1, D2, ...... is chosen. The decision-maker will specify the value of α depending upon his level of optimism. Example 4.2: Events

Act A1

A2

A3

E1

20

12

25

E2

25

15

30

E3

30

20

22

Solution: Let α = .6 for A1 max. pay-off = 30 Min. pay-off = 20 ∴ D1 = (.6 × 30 ) + (1 − .6 )20 = 26

Similarly, D2 D3

.6 30

.6 30 1 .6 22

1 .6 12 22.8 26.8

Since D3 is max, select the act A3.

4.2.3 Decision-Making under Risk Check Your Progress

In this situation, the decision-maker has to face several states of nature. But he has some knowledge or experience which will enable him to assign probability to the occurence of each state of nature. The objective is to optimize the expected profit, or to minimize the opportuiniy loss.

1. What are the two categories of decisions?

For decision problems under risk, the most popular methods used are Expected Monetary Value (EMV) criterion, Expected Opportunity Loss (EOL) criterion or Expected Value of Perfect Information (EVPI).

3. List the various choices available for the decisionmaker under uncertain situations.

2. What are events?


Operations Research

NOTES

(a) EMV Criterion When probabilities can be assigned to the various states of nature, it is possible to calculate the statistical expectation of gain for each course of action. The conditional value of each event in the pay-off table is multiplied by its probability and the product is summed up. The resulting number is the EMV for the act. The decision-maker then selects from the available alternative actions, the action that leads to the maximum expected gain (that is the action with highest EMV). Consider the following example. Let the states of nature be S1and S2 and the alternative strategies be A1 and A2. Let the pay-off table be as follows.

S1 S3

A1

A2

30 35

20 30

Let the probabilities for the states of nature S 1and S 2be 6 and 4 respectively . Then, EMV for A1 = (30 × .6) + (35 × .4) = 18 + 14 = 32 EMV for A2 = (20 × .6) + (30 × .4) = 12 + 12 = 24 Since EMV for A1 is greater, the decision-maker will choose the strategy A1. (b) EOL Criterion The difference between the greater pay-off and the actual pay-off is known as opportunity loss. Under this criterion, the strategy which has minimum Expected Opportunity Loss (EOL) is chosen. The calculation of EOL is similar to that of EMV. The following is an opportunity loss table. A1and A2 are the strategies and S1and S2 are the states of nature. A1

A2

S1

0

10

S2

2

–5

Let the probabilities for two states be .6 and .4. EOL for A1 = (0 × .6) + (2 × .4) = 0.8 EOL for A2 = (10 × .6) + (–5 × .4) = 6 – 2 = 4 EOL for A1 is least. Therefore, the strategy A1may be chosen. (c) EVPI Method The expected value of perfect information (EVPI) is the average (expected) return in the long run, if we have perfect information before a decision is to be made. In order to calculate EVPI, we choose the best alternative with the probability of their state of nature. EVPI is the expected outcome with perfect information minus the outcome wth max EMV. ∴ EVPI = Expected value with perfect information – max. EMV 138 Self-Instructional Material

Consider the following example.

Example 4.3: A1, A2, A3 are the acts and S1, S2, S3 are the states of nature. Also, known that P(S1) = .5, P(S2) = .4 and P(S3) = .1

Decision-Making

Solution: Pay-off table is as follows: Pay off table

State of nature

NOTES

A1

A2

A3

S1

30

25

22

S2

20

35

20

S3

40

30

35

EMV for A1 = (.5 × 30 ) + (.4 × 20 ) + (.1× 40 ) = 15 + 8 + 4 = 27 EMV for A 2 = (.5 × 25 ) + (.4 × 35) + (.1× 30) = 12.5 + 14 + 3 = 29.5 EMV for A 3 = (.5 × 22 ) + (.4 × 20) + (.1× 35) = 11 + 8 + 3.5 = 22.5 The highest EMV is for the strategy A2 and it is 29.5. Now to find EVPI, work out the expected value for maximum pay-off under all states of nature. Max. profit of each state

Probability

Expected value ( = Prob. × Profit)

S1

30

.5

15

S2

35

.4

14

S3

40

.1

4

∴ Expected pay-off with perfect information = 33. ∴ Thus, the expected value of perfect information (EVPI) = Expected Value with Perfect Information – Maximum EMV = 33 – 29.5 = 3.5. Example 4.4: You are given the following pay-offs of three acts A1, A2 and A3, and the states of nature S1, S2 and S3. States of nature A1 A2 A3 S1

25

–10

–125

S2

400

440

400

S3

650

740

750

The probabilities of the states of nature are respectively, .1, .7 and .2. Calculate and tabulate the EMV and conclude which of the acts can be chosen as the best. Solution: Act A1

Act A2

Act A3

Prob. × Pay off

Prob. × Pay off

Prob. × Pay off

.1 × –10 = –1

.1 × –125 = –12.5

.7 × 400 = 280

.7 × 440 = 308

.7 × 400 = 280

.2 × 650 = 130

.2 × 740 = 148

.2 × 750 = 150

.1 × 25 = 2.5

∴ EMV for A1 = 412.5, EMV for A2 = 455, EMV for A3 = 417.5 Since EMV is maximum for A2 choose the Act A2. Self-Instructional Material 139

Operations Research

NOTES

Example 4.5: A management is faced with the problem of choosing one of the products for manufacturing. The probability matrix after market research for the two products was as follows. State of nature Act Good Fair Poor Product A 0.75 0.15 0.10 Product B 0.60 0.30 0.10 The profit that the management can make for different levels of market acceptability of the products are as follows. State of nature Profit (in Rs) if market is Acts Good Fair Poor Product A 35,000 15,000 5,000 Product B 50,000 20,000 Loss of 3,000 Calculate the expected value of the choice of alternatives and advise the management. Solution: Let us put this information in a pay-off matrix with probabilities associated with the states of nature. Product A

Product B

Profit × Probability

Profit × Probability

35000 × 0.75 = 26250

50000 × 0.60 = 30000

Fair

1500 × 0.15 = 2250

20000 × 0.30 = 6000

Poor

5000 × 0.10 = 500

–3000 × 0.10 = –300

State of nature Good

EMV

29000

35700

Since the expected pay-off (EMV) of product B is greater, product B should be preferred by the management.

Preparation of pay-off table Example 4.6: An ink manufacturer produces a certain type of ink at a total average cost of Rs 3 per bottle and sells at a price of Rs 5 per bottle. The ink is produced over the weekend and is sold during the following week. According to past experience, the weekly demand has never been less than 78 or greater than 80 bottles in his place. You are required to formulate the pay-off table. Solution: The different states of nature are the demand for 78 units, 79 units or 80 units. Call them S1, S2, S3. The alternative courses of action are selling 78 units, 79 units or 80 units. Call them A1, A2, A3. Selling price of ink = Rs 5- per bottle Cost price = Rs 3- per bottle Calculation of pay-offs (Pay-off stands for the gain) Sale quantity × price – Production quantity × Cost 140 Self-Instructional Material

Decision-Making

A1S1 = 78 × 5 − 78 × 3 = 390 − 234 = 156 A 2S1 = 78 × 5 − 79 × 3 = 390 − 237 = 153 A 3S1 = 78 × 5 − 80 × 3 = 390 − 240 = 150 A1S 2 = 78 × 5 − 78 × 3 = 390 − 234 = 156

NOTES

A 2S 2 = 79 × 5 − 79 × 3 = 395 − 237 = 158 A 3S 2 = 78 × 5 − 80 × 3 = 395 − 240 = 155 A1S3 = 78 × 5 − 78 × 3 = 390 − 234 = 156 A 2S3 = 79 × 5 − 79 × 3 = 395 − 237 = 158 A 3S3 = 80 × 5 − 80 × 3 = 400 − 240 = 160

(Explanation: A1S1 means selling quantity is 78 and manufacturing quantity is 78. A2S1 means sales 78, production 79, and so on.) Pay-off Table Act (strategy)

State of nature (events)

A1

A2

A3

S1

156

153

150

S2

156

158

155

S3

156

158

160

Note: We shall show the states of nature in rows and acts in columns.

Preparation of loss table Example 4.7: An ink manufacturer produces a certain type of ink at a total average cost of Rs 3 per bottle and sells at a price of Rs 5 per bottle. The ink is produced over the weekend and is sold during the following week. According to past experience, the weekly demand has never been less than 78 or greater than 80 bottles in his place. You are required to formulate the loss table. Solution: Calculation of regret (opportunity loss) A1S1= 0 (since production and sales are of equal quantity say 78) A2S1 = 1× 3 = 3 (since one unit of production is in excess whose cost = Rs 3) A3S1 = 2 × 3 = 6 (since 2 units of production are in excess whose unit cost is @ Rs 3) A1S2 = 1 × 2 = 2 (since the demand of one unit is more than produced, the profit for one unit is Rs 2) Similarly, A2S2 = 0 (since units of production = units of demand) A1S3 = 2 × 2 = 4 A3S2 = 1 × 3 = 3 A2S3 = 2 × 1 = 2 and A3S3 = 0 Opportunity loss table State of nature (events)

Action A1

A2

A3

S1

0

3

6

S2

2

0

3

S3

4

2

0 Self-Instructional Material 141

Operations Research

NOTES

From the pay-off table oportunity loss table can also be prepared. Method: Let every row of the pay-off table represent a state of nature and every column represent a course of action. Then from each row select the highest pay-off and substract all pay-offs of that row from it. They are the opportunity losses. See the following examples. Example 4.8: The following is a pay-off table. From it, form a regret (opportunity loss) table. Pay-off table States of nature

A1

A2

A3

E1

156

153

150

E2

156

158

155

E3

156

158

160

Solution: Opportunity Loss Table Action

State of nature

A1

A2

A3

E1

156 –156 = 0

156 – 153 = 3

156 – 150 = 6

E2

158 – 156 = 2

158 – 158 = 0

158 – 155 = 3

E3

160 – 156 = 4

160 – 158 = 2

160 – 160 = 0

Example 4.9: The following is a pay-off table. Event (State of nature) Action E1 E2 E3 E4 A1 50 300 –150 50 400 0 100 0 A2 A3 –50 200 0 100 0 300 300 0 A4 Suppose that the probabilities of the events in this table are P(E1) = 0.15; P(E2) = 0.45; P(E3) = 0.25; P(E4) = 0.15. Calculate the expected pay-off. Prepare the opportunity loss table (Regret table) and calculate the expected loss of each action. Solution: Hint: Rewrite the question with E1, E2, E3, E4 as rows and A1, A2, A3, A4 as columns. Events E1 E2 E3 E4


A1 50 300 –150 50

Acts A2 400 0 100 0

A3 – 50 200 0 100

A4 0 300 300 0

Calculation of Expected Pay-off (EMV) A1 Pay-off × Prob. 50 × .15 = 7.5 300 × .45 = 135 –150 × .25 = –37.5 50 × .15 = 7.5 EMV = 112.5

A2 Pay-off × Prob. 400 × .15 = 60 0 × .45 = 0 100 × .25 = 25 0 × .15 = 0 EMV = 85

A3 Pay-off × Prob. – 50 × .15 = 75 200 × .45 = 90 0 × .25 = 0 100 × .15 = 15 EMV = 112.5

Decision-Making

A4 Pay-off × Prob. 0 × .15 = 0 300 × .45 = 135 300 × .25 = 75 0 × .15 = 0 EMV = 210

NOTES

Opportunity Loss Table E1 E2 E3 E4

A1 400 – 50 = 350 300 – 300 = 0 300 + 150 = 450 100 – 50 = 50

A2 400 – 400 = 0 300 – 0 = 300 300 – 100 = 200 100 – 0 = 100

A3 400 + 50 = 450 300 – 200 = 100 300 – 0 = 300 100 – 100 = 0

A4 400 – 0 300 – 300 300 – 300 100 – 0

= 400 =0 =0 =100

Calculation Expected Loss (EOL) A1 Loss × Prob. 350 × .15 = 52.5 0 × .45 = 0 450 × .25 = 112.5 50 × .15 = 7.5 EOL = 172.5

A2 Loss × Prob. 0 × .15 = 0 300 × .45 = 135 200 × .25 = 50 100 × .15 = 15 EOL = 200

A3 Loss × Prob. 450 × .15 = 67.5 100 × .45 = 45 300 × .25 = 75 0 × .15 = 0 EOL = 187.5

A4 Loss × Prob. 400 × .15 = 60 0 × .45 = 0 0 × .25 = 0 100 × .15 = 15 EOL = 75

Example 4.10: A newspaper boy has the following probability of selling a magazine: No. of copies sold Probability 10 0.10 11 0.15 12 0.20 13 0.25 14 0.30 The cost of a copy is 30 paise and sale price is 50 paise. He cannot return unsold copies. How many copies should he order? Solution: We can apply either EMV criterion or EOL criterion. Let us apply EMV criterion for which we have to calculate the pay-off. Number of copies ordered are the different courses of action. The copies ordered may be 10, 11, 12, 13, 14. Denote them by A1, A2, A3, A4, A5. Similarly, number of copies demanded may be 10, 11, 12, 13 or 14. These demands may be D1, D2, D3, D4, D5. These are events. The pay-off values are calculated as follows: Selling price of each item = 50 paise and cost of a copy = 30 paise A1D1 = (10 × 50) – (10 × 30) = 200 A2D1 = (10 × 50) – (11 × 30) = 170 A3D1 = (10 × 50) – (12 × 30) = 140 A4D1 = (10 × 50) – (13 × 30) = 110, and so on. Self-Instructional Material 143

Operations Research

The 25 pay-off values can thus be calculated. They are shown as follows: D1 D2 D3 D4 D5

NOTES

A1 200 200 200 200 200

A2 170 220 220 220 220

A3 140 190 240 240 240

A4 110 160 210 260 260

A5 80 130 180 230 280

The given probabilities are .10, .15, .20, .25, .30. Calculation of EMV for all the acts. A1 Pay off × Prob. .10 × 200 = 20 .15 × 200 = 30 .20 × 200 = 40 .25 × 200 = 50 .30 × 200 = 50 200

A1 Pay off × Prob. .10 × 170 = 17 .15 × 220 = 33 .20 × 220 = 44 .25 × 220 = 55 .30 × 220 = 66 215

A4 Pay off × Prob .10 × 110 = 20 .15 × 160 = 30 .20 × 210 = 40 .25 × 260 = 50 .30 × 260 = 50 220

A1 Pay off × Prob. .10 × 140 = 20 .15 × 190 = 28.5 .20 × 240 = 48 .25 × 240 = 60 .30 × 240 = 72 222.5

A5 Pay off × Prob .10 × 80 = 8 .15 × 130 = 19.5 .20 × 180 = 36 .25 × 230 = 55 .30 × 280 = 84 205

∴ EMV for the acts A1, A2, A3, A4, and A5 are respectively 200, 215, 222.5, 220 and 205. EMV for A3 is greater and therefore A3 is optimal act. ∴ No. of copies to be ordered = 12. Example 4.11: A grocery store with a bakery department is faced with the problem of how many cakes to buy in order to meet the day’s demand. The grocer prefers not to sell day-old goods in competition with fresh products. Leftover cakes are, therefore, a complete loss. On the other hand, if a customer desires a cake and all of them have been sold, the disappointed customer will buy elsewhere and the sales will be lost. The grocer has, therefore collected information on the past sales in a selected 100-day period as shown in the following table: Sales per day

No. of days

Probability

25 26 27 28

10 30 50 10

0.10 0.30 0.50 0.10

100

1.00

Construct the pay-off table and the opportunity loss table. What is the optimal number of cakes that should be bought each day? Apply both EMV and EOL criteria. 144 Self-Instructional Material

Also find (and interpret) EVPI (Expected Value of Perfect Information). A cake costs Rs 0.80 and sells for Re 1. Solution: Let A1, A2, A3, and A4 for strategies and S1, S2, S3, S4 stands for states of nature. Then A1, A2, A3, A4 respectively stand for stocking 25, 26, 27,28 cakes. S1, S2, S3, S4 respectively stand for demands for 25, 26, 27, 28 cakes.

Decision-Making

NOTES

Conditional pay-off values can be obtained as explained earlier. The pay-off values thus obtained are as follows. Pay-off table State of nature (Demand) S1(25) S2(26) S3(27) S4(28)

A1(25) 5.00 5.00 5.00 5.00

Alternative strategies A2(26) A3(27) 3.40 4.20 4.40 5.20 5.40 5.20 5.40 5.20

A4(28) 2.60 3.60 4.60 5.60

Probabilities are .10, .30, .50, .10 EMV for Act A1 = (5 × .10) + (5 × 30) + (5 × .50) + (5 × .10) = .5 Similarly, EMV values A2, A3, A4 are 5.10, 4.9, 4.2 respectively. The maximum EMV is seen strategy A2. Thus, according to the EMV decision criterion, the store would stock 26 cakes (A2). Regret (Opportunity loss) table State of nature (Demand) S1(25) S2(26) S3(27) S4(28)

A1(25) 0 0.2 0.4 0.6

Alternative strategies A2(26) A3(27) 1.6 0.8 0.8 0 0 0.2 0.2 0.4

A4(28) 2.4 1.6 0.8 0

Probabilities are 0.10, 0.30, 0.50, 0.10 Expected opportunity loss for A1 = (0 × .10 ) + (.2 × .30) + (.4 × .50 ) + (.6 × .10 ) = .32. Similarly for A2, A3, A4 expected opportunity losses are 0.22, 0.42 and 1.12 respectively. EOL is least for A2. \ A2 is the optimal Act. Hence, 26 is the optimal number to buy. To find EVPI, select the highest pay-off in each row and find the expected value. Then we have Pay-off Prob. Payoff × Prob. 5.00 0.10 0.50 5.20 0.30 1.56 5.40 0.50 2.70 5.60 0.10 0.56 Total 5.32 Self-Instructional Material 145

Operations Research

∴ Expected value with perfect information = 5.32

Max. EMV is for Act A2 which is equal to 5.10.

NOTES

Expected value of perfect information (EVPI) = Expected Value with Perfect Information – Highest EMV (EMV of A2) = 5.32 – 5.10 = .32 Example 4.12: A factory produces 3 varieties of fountain pens. The fixed and variable costs are as follows: Fixed Cost Variable Cost Type 1 Rs 2,00,000 Rs 10 Type 2 Rs 3,20,000 Rs 8 Type 3 Rs 6,00,000 Rs 6 The likely demands under three situations are given as follows: Demand Units Poor 25,000 Moderate 1,00,000 High 1,50,000 If the price of each type is Rs 20, prepare the pay-off table after showing necessary calculations. Solution: Let T1, T2, T3 stand for Type 1, Type 2, and Type 3 and D1, D2 and D3 for poor demand, moderate demand and high demand respectively. The pay-off (in thousands) = Sales revenue from the estimated demand – Total variable cost – Fixed cost T1 D1 = (20 × 25) − (10 × 25) − 200 = 500 − 250 − 200 = +50 T2 D1 = (20 × 25) − (8 × 25) − 320 = 500 − 200 − 320 = −20 T3 D1 = (20 × 25) − (6 × 25) − 600 = 500 − 150 − 600 = −250 T1 D2 = (20 × 100) − (10 × 100) − 200 = 2000 − 1000 − 200 = +800 T2 D2 = (20 × 100) − (8 × 100) − 320 = 2000 − 800 − 320 = +880 T3 D2 = (20 × 100) − (6 × 100) − 600 = 2000 − 600 − 600 = +800 T1 D3 = (20 × 150) − (10 × 150) − 200 = 3000 − 1500 − 200 = +1300 T2 D3 = (20 × 150) − (8 × 150) − 320 = 3000 − 1200 − 320 = +1480 T3 D3 = (20 × 150) − (6 × 150) − 600 = 3000 − 900 − 600 = +1500

The pay-off table (in ’000s) D1 D2 D3

Note: (i)

T1 50 800 1300

T2 –20 880 1480

T3 –250 800 1500

Expected value of sales = Sum of the products of various values of sales and probabilities (ii) Expected Monetary value of an act = Sum of the products of various values of the pay off of the act and probabilities (iii) Expected value of cost = Sum of the products of the various values of the cost and probabilities (iv) Expected value of loss of an act = Sum of the products of the losses of the acts and Probabilities


Example 4.13: RM (Mudland) Limited has recently installed new machinery but has not yet decided on the appropriate number of a certain spare part required for repairs. Spare parts cost Rs 2,000 each but are only available if ordered now. If the plant failed and there was no spare part available, the cost to the business of mending the plant rises to Rs 15,000. The plant has an estimated life of ten years and the probability distribution of failure during this time, based on the experience with similar plant, is as follows. Number of failure over ten-year period Probability 0 1 2 3 4 5 and over

Decision-Making

NOTES

0.1 0.4 0.3 0.1 0.1 Nil

(For the purpose of this question, ignore any discounting for time value of money) You are required to calculate: (a) The expected number of failures in the ten-year periods (b) The optimal number of spares that should be purchased now: (Include in your workings a table to show the cost of alternative ordering policies and failures) (c) The current cost of the ordering policy chosen; (d) The value of perfect information of the number of failures in the tenyear life. Solution: (a) Expected number of failures = (0 × 0.1) + (1× 0.4 ) + (2 × 0.3) + (3 × 0.1) + (4 × 0.1) = 1.7

(b) Pay-off values are calculated as follows: State of nature (Number of failure) E1(0) E2(1) E3(2) E4(3) E5(4)

Alternative strategies (Spare parts purchased) (in ‘000) A2 A3 A4 A5 A1 (0) (1) (2) (3) (4) 0 2 4 6 8 15 2 4 6 8 30 17 4 6 8 45 32 19 6 8 60 47 34 21 8

Explanation for calculation of pay-off values: Calculation of A1E1 = (0 × 2000) + ( 0) 15000 = 0 A1E2 = (0 × 2000) + (1 – 0) 15000 = 15000 A1E2 = (2 × 2000) + (0 × 15000) = 4000 A3E4 = (2 × 2000) + (3 – 2) 15000 = 19000, etc. Self-Instructional Material 147

Operations Research

NOTES

EMV for each A1, A2, A3, A4 and A5 are 25.5, 14, 8.5, 7.5, 8 respectively selecting the act with least expected value, Optimal Act is A3. ∴ Purchase three parts (c) If three spare parts are purchased, the expected cost is a minimum and the cost current of the ordering policy chosen = 3 × 2,000 = Rs 6000. (d) If one had perfect information, in advance, of the number of failures that would occur one would purchase the corresponding number of spare parts and expected cost under perfecet information (ECPI). (0.1 0) (0.4 2000) (0.3 4000) (0.1 6000) (0.1 8000)

Rs 3400

The expected value of perfect information = Optimal EMV – ECPI = Rs 7500 – Rs 3,400 = 4,100. Example 4.14: A food products compay is counterplanning the introduction of a revolutionary new product with new packing to replace the existing product at much higher price (S1) or a moderate change in the composition of the existing product with a new packaging at a small increase in price (S2) or a small change in the composition of the existing expect the word, ‘New with a negligible increase in the price (S3). The three possible states of nature of events are: (i) high increase in sales (N1). (ii) no change in sales (N2) and (iii) decrease is sales (N3). The marketing department of the company worked out the pay-offs in terms of yearly new profits for each of the strategies or these events (expected sales). This is represented in the following table. Pay-offs State of nature N2 N3 Strategies N1 S1 700 300 150 500 450 0 S2 S3 300 300 300 Which strategy should the executive concerned choose on the basis of: (a) Maximum criterion (b) Maximax criterion (c) Minimax regret criterion (d) Laplace criterion? Solution: Writing the pay-off table properly State of nature N1 N2 N3

S1 700 300 150

Strategies S2 500 450 0

S3 300 300 300

(a) Maximin criterion S1 S2 S3

Minimum pay offs 150 0 300

Max of these Minima = 300 The executive should choose strategy S3. 148 Self-Instructional Material

(b) Maximax criterion

Decision-Making

Maximum pay-offs 700 500 300

S1 S2 S3

NOTES

Maximum of maxima = 700 ∴ The executive can choose Act S1. (c) Minimax regret criterion Opportunity loss table N1 N2 N3

S1 700 – 700 = 0 450 – 300 = 150 300 – 150 = 150

Maximum for S1 for S2 for S3

S2 700 – 500 = 200 450 – 450 = 0 300 – 0 = 300

S3 700 – 300 = 400 450 – 300 = 150 300 – 300 = 0

Opportunity loss – 150 – 300 – 400

The executive should choose strategy S1 since it minimizes the maximum of the losses. (d) Laplace criterion Assigning equal probability (say 1/3) to each state of nature calculate the expected monetary value for each act. Prob. × Pay-off: ⎞ 1150 ⎞ ⎛1 ⎛1 ⎞ ⎛1 = 383.33 ⎜ × 700 ⎟ + ⎜ × 300 ⎟ + ⎜ × 150 ⎟ = 3 ⎠ ⎠ ⎝3 ⎝3 ⎠ ⎝3 ⎞ ⎛ 1 ⎞ 950 ⎞ ⎛1 ⎛1 = 316.666 ⎜ × 500 ⎟ + ⎜ × 450 ⎟ + ⎜ × 0 ⎟ = 3 ⎠ ⎝3 ⎠ ⎠ ⎝3 ⎝3 ⎞ 900 ⎞ ⎛1 ⎞ ⎛1 ⎛1 = 300 ⎜ × 300 ⎟ + ⎜ × 300 ⎟ + ⎜ × 300 ⎟ = 3 3 3 3 ⎠ ⎠ ⎝ ⎠ ⎝ ⎝

Since the EMV is highest for strategy 1, the executive should select strategy S1. Example 4.15: The research department of consumer products division has recommended to the marketing department to launch a soap with three different perfumes. The marketing manager has to decide the type of perfume to launch under the following estimated pay-off for the various levels of sales. Estimated level of sales (units) Types of perfume 20,000 I 250 II 40 III 60

10,000 15 20 25

2,000 10 5 3

Examine which type can be chosen under maximax, minimax, maximin and Laplace criterion.


Operations Research

Solution: Rewriting the pay off table Levels of Sales

NOTES

Types of perfume

20000

I 250

II 40

III 60

10000

15

20

25

2000

10

5

3

(1) Maximax criterion Maximum for Type I = 250 Maximum for Type II = 40 Maximum for Type III = 60 Maximum of maximum = 250 ∴ Select type I perfume (2) Minimax criterion Loss table is Sales 20,000 10,000 2,000

I 0 10 0

Type II 210 5 5

III 190 0 7

Maximum losses under I, II, III are respectively 10, 210, 190 (from the pay-off table) Minimum of these is 10. ∴ Type I is preferred. (3) Maximin criterion Minimum pay-off under each type I, II, III are respectively 10, 5, 3 (from the pay-off table) Maximum of these is 10. ∴ Type I is preferred. (4) Laplace criterion Let the probability for each levels of sales be taken as 1/3 each. Expected pay-offs are: 1⎞ ⎛ 1⎞ ⎛ 1 ⎞ 275 ⎛ Type I → ⎜ 250 × ⎟ + ⎜15 × ⎟ + ⎜10 × ⎟ = = 91.67 3 3 3⎠ 3 ⎠ ⎝ ⎠ ⎝ ⎝ 1⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ 65 ⎛ Type II → ⎜ 40 × ⎟ + ⎜ 20 × ⎟ + ⎜ 5 × ⎟ = = 21.67 3⎠ ⎝ 3⎠ ⎝ 3⎠ 3 ⎝ 1⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ 88 ⎛ Type III → ⎜ 60 × ⎟ + ⎜ 25 × ⎟ + ⎜ 3 × ⎟ = = 29.33 3 3⎠ ⎝ 3⎠ 3 ⎝ ⎠ ⎝

Maximum expected pay-off is for Type I, so choose Type I perfume. 150 Self-Instructional Material

Example 4.16: A company has an opportunity to computerize its records department. However, the existing personnel have job security under union agreement. The cost of the three alternative programmes for the changeover depend upon the attitude of the union and is estimated as follows: Attitude of union General retraining Selective retraining Hire new employees Antagonist

940

920

900

Passive

810

800

820

Enthusiastic

700

710

860

Decision-Making

NOTES

(i) Construct the opportunity losses table. Solution: Opportunity loss for row I is obtained by subtracting 900 (least element of row I) from all the elements of I row are 940 – 900 = 40,

920 – 900 = 20,

900 – 900 = 0

Similarly, for row II the opportunity losses are 810 – 800 =10, 800 – 800 = 0 and 820 – 800 = 20 Similarly for row III, they are 700–700 = 0,

710–700 = 10,

860-700 = 160

Thus, the opportunity loss table is Attitude of the nation Antagonist Passive Enthusiastic

4.3

State of action General Selective Hire new 40 20 0 10 0 20 0 10 160

DECISION TREE ANALYSIS

A ‘Decision tree’ is one of the tools used for the diagramatic presentation of the sequential and multidimensional aspects of a particular decision problem for systematic analysis and evaluation. Here, the decision problem, the alternative courses of action, the states of nature and the likely outcomes of alternatives are diagramatically or graphically depicted as branches and sub-branches of a horizontal tree. The decision tree consists of nodes and branches. The nodes are of two types, decision nodes and chance nodes. Courses of action (or strategies) originate from the decision nodes as the main branches. At the terminal of each main branch there are chance nodes. From these chance nodes, chance events emanate in the form of sub-branches. The respective pay-offs and probabilities associated with alternative courses and chances events are shown along the sub-branches. At the terminal of the sub-branches are shown the expected values of the outcome.


Operations Research

NOTES

Fig 4.1 Decision Tree

Here A1, A2, A3 and A4 are strategies E1, E2, E3 are events O11, O12, O21, O22, O31, O32 are outcomes. A decision tree is highly useful to a decision-maker in multistage situations which involve a series of decisions each dependent on the preceding one. Working backward, from the future to the present, we are able to eliminate unprofitable branches and determine optimum decisions. The decision tree analysis allows one to understand, simply by inspection, various assumptions and alternatives in a graphic form which is much more easier to understand than the abstract analytical form. The advantages of the decision tree structure are that complex managerial problems and decisions of a chain-like nature can be systematically and explicitly defined and evaluated. Example 4.17: A firm owner is seriously considering to drill a farm well. In the past, only 70% of wells drilled were successful at 200 feet of depth in the area. Moreover, on finding no water at 200 ft., some persons drilled it further up to 250 feet but only 20% struck water at 250ft. The prevailing cost of driling is Rs 50 per foot. The farm owner has estimated that in case he does not get his own wells, he will have to pay Rs 15,000 over the next 10 years, in present value (PV) term, to buy water from the neighbour. The following decisions can be optimal. (i) Do not drill any well (ii) drill up to 200ft and (iii) if no water is found at 200 ft. drill further up to 250ft. Draw an appropriate decision tree and determine the farm owner’s strategy under EMV approach. Solution:

Decision Tree for the Firm Owner

At D2 point


Decision

:

(a) drill upto 250feet (b) Do not drill

Event

:

(a) No water

(b) Water

Probabilities are .2, .8

Decision-Making

EMV for drill up to 250 feet. = (12500 × .2) + (27500 × .8) = 24500 EMV for do not drill = 25000 (from the tree)

NOTES

EMV is smaller for the act drill upto 250 feet. So it is an optimal act. At D2 point The decisions are drill upto 200 feet and do not drill. Events are same as those of D2 point. Probabilities are .7, .3 EMV for drill up to 200 feet = (10000 × .7) + (24500 × .3) = 14350 EMV for do not drill = 15,000 from the tree. The optimal decision is drill upto 200 feet (as the EMV is small). Therefore, combining D1and D2 the optimal strategy is to drill the well upto 200 feet and if no water is struck, then further drill it to 250 feet. Example 4.18: A firm is planning to develop and market a new drug. The cost of extensive research to develop the drug has been estimated at Rs. 100000. The manager of the research programme has found that there is a 60% chance that the drug will be developed successfully. The market potential has been assessed as follows: Market condition

Prob.

Present value of profit (Rs)

Large Market potential

0.1

50,000

Moderate Market potential

0.6

25,000

Low Market potential

0.3

10,000

The present value figures do not include the cost of research. While the firm is considering this proposal, a second proposal almost similar comes up for consideration. The second one also requires an investment of Rs 1,00,000 but the present value of all profits is Rs 12,000. The return on investment in the second proposal is certain. (i) Draw a decision tree indicating all events and choices of the firm. (ii) What decision should the firm take regarding the investment of Rs 1,00,000? Solution:

Decision Tree for the Drug Firm


Operations Research

NOTES

At point D2 Decisions are (1) Enter market (2) Do not enter market. (a) Enter market EMV = Expected P V = (50000 × .1) + (25000 × .6) + (10000 ×.3) = 5000 + 15000 + 3000 = 23000 (b) Do not enter market EMV = Expected PV = 0 × 1 = 0 Decision: Enter the market since EMV is more. At point D1 Decisions are (1) Develop new drug (2) Accept proposal II (a) Develop new drug EMV = Expected PV = (23000 × .6) + (0 × .4) = (13800 + 0) =13800 (b) Accept Proposal II EMV = Expected PV = 12000 × 1 = 12000 Using EMV criterion, the optimal decision at D1 is to develop and market the new drug.

4.4

INTEGER PROGRAMMING AND DYNAMIC PROGRAMMING: CONCEPTS AND ADVANTAGES

A linear programming problem in which all or some of the decision variables are constrained to assume non-negative integer values is called an Integer Programming Problem. (IPP). In a linear programming problem if all variables are required to take integral values then it is called the pure (all) integer programming problem (Pure IPP). If only some of the variables in the optimal solution of an IPP are restricted to assume non-negative integer values while the remaining variables are free to take any non-negative values, then it is called a mixed integer programming problem (Mixed IPP). Further, if all the variables in the optimal solution are allowed to take values 0 or 1, then the problem is called the 0–1 programming problem or standard discrete programming problem. The general integer programming problem is given by Max Z = CX. Subject to the constraints Ax 0 and some or all variables are integers


4.4.1 Importance of Integer Programming Problems In IPP, all the decision variables were allowed to take any non-negative real values as it is quite possible and appropriate to have fractional values in many situations. There are several frequently occurring circumstances in business and industry that lead to planning models involving integer valued variables. For example, in production, manufacturing is frequently scheduled in terms of batches, lots or runs. In allocation of goods, a shipment must involve a discrete number of trucks and aircrafts. In such cases, the fractional values of variables like 13/3 may be meaningless in the context of the actual decision problem.

Decision-Making

NOTES

This is the main reason why integer programming is so important for marginal decisions.

4.4.2 Applications of Integer Programming Integer programming is applied in business and industry. All assignment and transportation problems are integer programming problems, as in assignment and travelling salesmen problem, all the decision variables are either zero or one. i.e., xij = 0 or 1 Other examples are capital budgeting and production scheduling problems. In fact, any situation involving decisions of the type ‘either to do a job or not to do’ can be viewed as an IPP In all such situations xij = 1 if the jth activity is performed, 0 if the jth activity is not performed. In addition, allocation problems involving the allocation of men, and machines give rise to IPP, since such communities can be assigned only in integers and not in fractions. Note: If the non-integer variable is rounded off, it violates the feasibility and also there is no guarantee that the rounded off solution will be optimal. Due to these difficulties, there is a need for developing a systematic and efficient procedure for obtaining the exact optimal integer solution to such problems. Many decision-making problems involve a process that takes place in such a way that at each stage, the process is dependent on the strategy chosen. Such type of problems are called Dynamic Programming Problem (DPP). Thus, DPP is concerned with the theory of multistage decision process. Mathematically, a DPP is a decision making problem in n variables, the problem being subdivided into n sub-problems (segments), each sub-problem being a decision-making problem in one variable only. The solution to a DPP is achieved sequentially, starting from one (initial) stage to the next, till the final stage is reached.

4.4.3 Decision Tree and Bellman’s Principle of Optimality A multistage decision system in which each decision and state variable can take only finite number of values can be represented graphically by a decision tree.


Operations Research

Stage 1

Stage 2

NOTES

Stage 3

Circles represent nodes corresponding to stages and lines between circles denote arcs corresponding to decisions. The dynamic programming technique deals with such situations by dividing the given problem into sub-problems or stages. Bellman’s principle of optimality states that ‘An optimal policy (set of decisions) has the property that whatever the initial state and decisions are, the remaining decisions must constitute an optimal policy with regard to the state resulting from the first decision.’ The problem which does not satisfy the principle of optimality cannot be solved by the dynamic programming method.

4.4.4 Characteristics of DPP The basic features which characterize the dynamic programming problem are as follows.

Check Your Progress 4. Mention any two methods which are used during decision-making under risk. 5. Define a decision tree. 6. State an advantage of decision tree analysis. 156 Self-Instructional Material

1. The problem can be divided into stages with a policy decision required at each stage. 2. Every stage consists of a number of states associated with it. The states are the different possible conditions in which the system may find itself at that stage of the problem. 3. Decision at each stage converts the current stage into the state associated with the next stage. 4. The state of the system at a stage is described by a set of variables called state variables. 5. When the current state is known, an optimal policy for the remaining stages is independent of the policy of the previous ones. 6. The solution procedure begins by finding the optimal policy for each state to the last stage. 7. A recursive relationship which identifies the optimal policy for each state with n stages remaining, given the optimal policy for each state with (n[1) stages left. 8. Using recursive equation approach, each time the solution procedure moves backward stage by stage for obtaining the optimum policy of each state for the particular stage, till it attains the optimum policy beginning at the initial stage. Note: A stage may be defined as the portion of the problem that possesses a set of mutually exclusive alternatives from which the best alternative is to be selected.

4.4.5 Dynamic Programming Algorithm

Decision-Making

The solution of a multistage problem by dynamic programming involves the following steps. Step 1: Identify the decision variables and specify the objective function to be optimized under certain limitations, if any.

NOTES

Step 2: Decompose the given problem into a number of smaller sub-problems. Identify the state variable at each stage. Step 3: Write down the general recursive relationship for computing the optimal policy. Decide whether forward or backward method is to be followed to solve the problem. Step 4: Construct appropriate stages to show the required values of the return function at each stage. Step 5: Determine the overall optimal policy or decisions and its value at each stage. There may be more than such optimal policy.

4.4.6 Advantages of DPP 1. Generally the solution of a recursive equation involves two types of computations depending upon whether the system is continuous or discrete. In the first case, the optimal decision at each stage is obtained by using the usual classical method of optimization. In the second case, a tabular computational scheme is followed. 2. The D.P.P is solved by using the recursive equation, starting from the first to the last stage i.e. obtaining the sequence f1 f2 f3 ...fn of the optimal solution. This computation is called the forward computational procedure. If the recursive equations are formulated to obtain a sequence fn fn–1 ...f1 then the computation is known as the backward computational procedure. Example 4.19: Use the principle of optimality to find the Maximum value of Z = b1x1 + b2x2 + ... + bnxn when

x1 + x2 + x3 + ... + xn = C x1x2 ... xn ≥ 0

Solution: The problem can be considered to divide the positive quantity C in n parts x1x2 ... xn so that b1x1 + b2x2 + ... + bnxn is maximum. Let fn (C) = b1x1 + b2x2 + ... + bnxn .

Recursive equation If Zi be the ith part (i = 1, 2 ... n) of the quantity then the recursive equation of the problem are f1(x1) = max b1Z1 = b1x1 Z1 = x1 and

fi(xi) = max [biZi + fi – 1 (xi – Zi)]

0 ≤ Zi ≤ xi[where i = 1, 2 ... n] Self-Instructional Material 157

Operations Research

Solution of Recursive Equations For one stage problem i = 1

NOTES

f1(x1) = b1x1 This gives, f1 (C ) = b1C which is trivially true. For two stage problem, where i = 1, 2 f2(x2) = max (b2Z2 + f1(x2 – Z2) 0 ≤ Z2 ≤ x2 f2(C) = max (b2Z2 + f1(C – Z) 0 ≤ Z2 ≤ C for x2 = C1, Z2 = Z = max b2Z + b1(C – Z) 0≤Z≤C = max (b2 – b1) Z + b1C 0≤Z≤C If b2 – b1 is positive then this is maximum for Z = C, otherwise, it will be minimum. Thus f 2 (C ) = b2C Similarly for three stage problem ( i = 1, 2 3) f3(x3) = max [(b3Z3 + f2(x3 – Z3)] 0 ≤ Z3 ≤ x3 f3(C) = max [(b3Z + f2(C – Z)[ 0 ≤ Z3 ≤ C = max [b3Z + b2(C – Z)] 0≤Z≤C = max (b3 – b2) C + b2C 0≤Z≤C Again if b3 – b2 is positive, then it gives maximum value for Z = C otherwise, it gives the minimum value. Thus f 2 (C ) = b3C From the results of the three stages 1, 2, 3 it can be easily shown by induction method that f n (C ) = bn C

Hence, the optimal policy will be (0, 0, 0 ... xn = C) with f1(C) = bnC


Example 4.20: Use dynamic programming to show that

Decision-Making

Z = p1 log p1 + p2 log p2 + .... + pn log pn Subject to the constraints p1 + p2 + .... + pn = 1 and Pj ≥ 0 (j = 1, 2 .. n) is minimum, where p1 = p2 =... pn =

1 n

NOTES

Solution. The problem here is to divide unity into n parts so as to minimize the quantity Σpi log pi Let fn(1) denote the minimum attainable sum of p1 log p1 (i = 1, 2, .... n) For n = 1 (Stage 1) f1(1) = min (p1 log p1) = 1 log 1 0
1 satisfying the condition 0 < x ≤ 1 2 1 1 ⎛ 1⎞ ⎛ 1⎞ 1⎞ ⎛1 log + ⎜ 1 – ⎟ log ⎜ 1 – ⎟ = 2 ⎜ log ⎟ 2 2 ⎝ 2⎠ ⎝ 2⎠ 2⎠ ⎝2

Similarly for stage 3, the minimum value of the recusive equation is obtained as f3(1) = min [x log x + f2 (1 – x)] 0≤x≤1 Self-Instructional Material 159

Operations Research

NOTES

⎛1− x ⎞ ⎛1– x ⎞ = min [x log x + 2 ⎜ ⎟ log ⎜ ⎟] ⎝ 2 ⎠ ⎝ 2 ⎠

0≤x≤1 Now, since the minimum value of ⎛1− x ⎞ ⎛1– x ⎞ x log x + 2 ⎜ ⎟ log ⎜ ⎟ is ⎝ 2 ⎠ ⎝ 2 ⎠

attained at x =

1 satisfying 0 < x £ 1 we have 3

f3(1) = ∴ optimal policy is P1 = P2 = P3 =

1 1 1⎞ 1⎞ ⎛1 ⎛1 log + 2 ⎜ log ⎟ = 3 ⎜ log ⎟ 3 3 3⎠ 3⎠ ⎝3 ⎝3 1 3

In general for n stage problem, we assume the optimal policy to be P1 = P2 = ... Pn =

1 and fn(1) = n n

1⎤ ⎡1 ⎢⎣ n log n ⎥⎦

This can be shown easily using mathematical induction. For n = m + 1, the recursive equation is fm + 1 (1) = min [x log x + fm(1 – x)] 0≤x≤1 ⎡ ⎛ 1 − x ⎞ ⎛ 1 − x ⎞⎤ = min ⎢ x log x + m ⎜ ⎟ log ⎜ ⎟⎥ ⎝ m ⎠ ⎝ m ⎠⎦ ⎣

0≤x≤1 =

1 1 1 ⎞ ⎛ 1 log log + m⎜ ⎟ m +1 m +1 m +1⎠ ⎝ m +1

1 ⎞ ⎛ 1 log =m+1 ⎜ ⎟ m +1⎠ ⎝ m +1

Since minimum x log x +

1 1− x 1− x is attained at x = . log m +1 x m

1⎞ 1⎞ ⎛1 ⎛1 1 Hence, the required optimum policy is ⎜ , .... ⎟ with f n (1) = n ⎜ log ⎟ n⎠ n⎠ ⎝n ⎝n n


4.5

SUMMARY

In this unit, you have learned about decision-making and its applications. In any decision problem, the decision-maker is concerned with choosing the method which yields the best result from among the various alternative courses of action. The unit has explained that when the decision-maker faces multiple states of nature and has no means to arrive at probability values of the likelihood of occurrence of these states of nature, the problem is a decision problem under uncertainty. You have also learned that during decision-making under risk, the decision-maker has some knowledge which enables him to assign probability to the occurrence of each state of nature. You have also learned how to prepare a pay-off table. It has also described decision tree analysis, where you have learned that the decision problem, alternative courses of action, states of nature and the likely outcomes of alternatives are depicted as if they are branches and sub-branches of a horizontal tree. The unit has also explained the characteristics of integer programming and dynamic programming.

4.6

Decision-Making

NOTES

KEY TERMS

●

Tactical decisions: Decisions which affect a business in the short run.

●

Strategic decisions: Decisions which have far-reaching effects on the course of business.

●

Pay-off table: It represents the economics of a problem, i.e., revenue and costs associated with any action with a particular outcome. It is an ordered statement of profit or costs resulting under the given situation.

●

Opportunity loss table: It is the loss incurred because of failure to take the best possible action. Opportunity losses are calculated separately for each state of nature that might occur.

●

Decision-making under certainty: In this case, the decision-maker knows with certainty the consequences of every alternative or decision choice. The decision-maker presumes that only one state of nature is relevant for his purpose.

●

Decision-making under uncertainty:When the decision-maker faces multiple states of nature but he has no means to arrive at probability values to the likelihood of occurence of these states of nature, the problem is a decision problem under uncertainty.

●

Decision tree: A diagrammatic presentation of the sequential and multidimensional aspects of a particular decision problem for systematic analysis and evaluation.

●

Pure IPP: In a linear programming problem if all variables are required to take integral values then it is called the pure (all) integer programming problem (Pure IPP).

●

Mixed IPP: If only some variables in the optimal solution of an IPP are restricted to assume non-negative integer values while the remaining variables are free to take any non-negative values, then it is called a mixed integer programming problem (Mixed IPP). Self-Instructional Material 161

Operations Research

4.7

ANSWERS TO ‘CHECK YOUR PROGRESS’ 1. Tactical and strategic.

NOTES

2. Events are the occurrences which affect the achievement of objectives. They are also called states of nature or outcomes. 3. (i) Maximax criterion, (ii) Minimax criterion, (iii) Maximin criterion, (iv) Laplace criterion and (v) Hurwicz alpha criterion 4. Expected Monetary Value criterion and Expected Opportunity Loss criterion. 5. A decision tree is a diagrammatic presentation of the sequential and multidimensional aspects of a particular decision problem for systematic analysis and evaluation. 6. Complex managerial problems can be systematically and explicitly defined and evaluated.

4.8

QUESTIONS AND EXERCISES

Short-Answer Questions 1. What do you understand by ‘decision theory’? 2. What are EMV and EOL criteria? 3. What are pay-off tables and regret tables?

Long-Answer Questions 1. Describe some methods which are useful for decision-making under uncertainty. 2. Explain the terms (i) Expected Monetary Value (ii) Expected Value of Perfect Information. 3. Write notes on the Bayesian decision theory. 4. Write a note on decision tree. 5. Write a note on action space. 6. Explain (a) Maximax (b) Minimax (c) Maximin decision criteria. 7. From the following pay-off table, decide which is the optimal act. Strategies Events A1 A2 A3 S1 20 40 60 15 –10 –15 S2 35 25 –20 S3 P(S2) = .5 P(S3) = .1 P(S1) = .4 EMV for A1 = 19, for A2 = 13.5, for A3 = 13.5 So A1] 162 Self-Instructional Material

4.9

FURTHER READING

Bierman, Harold, Charles Bonini and W.H. Hausman. 1973. Quantitative Analysis for Business Decisions. Homewood (Illinois): Richard D. Irwin.

Decision-Making

NOTES

Gillett, Billy E. 2007. Introduction to Operations Research. New Delhi: Tata McGraw-Hill. Lindsay, Franklin A. 1958. New Techniques for Management Decision Making. New York: McGraw-Hill. Herbert, G. Hicks. 1967. The Management of Organisations. New York: McGrawHill. Massie, Joseph L. 1986. Essentials of Management. New Jersey: Prentice-Hall. Ackoff, R.L. and M.W. Sasieni. 1968. Fundamentals of Operations Research. New York: John Wiley & Sons Inc. Enrick, Norbert Lloyd. 1965. Management Operations Research. New York: Holt, Rinehart & Winston.


BCA-504

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